Answer:
if you meant %, the answer is 2016
Step-by-step explanation:
1120 + 80% is 2016
HELP ME PLSS OMLLL IMA CRY
Answer:
1. 15 pieces of candy
2. 48 cows
3. 81 ham subs
4. 24 boys
Step-by-step explanation:
Assuming you just want answers, but lmk
Find the difference between the product of 2.5 and 7.5and the sum of 2.7 and 9.55
Two numbers have a sum of 35 and a product of 250. what are the numbers
Answer:
25 and 10
Step-by-step explanation:
because 25+10=35 and 25*10=250
1 (Express 360° into centesimal system.)
Answer: 400 gradians
Reason:
100 gradians = 90 degrees
4*100 gradians = 4*90 degrees
400 gradians = 360 degrees
Kadeesha invested
$
900
$900 in an account that pays 1.5% interest compounded annually. Assuming no deposits or withdrawals are made, find how much money Kadeesha would have in the account 11 years after her initial investment. Round to the nearest tenth (if necessary).
The amount of money Kadeesha have in her account after 11 years is $1059.3.
What is the compound interest?Compound interest is the interest on savings calculated on both the initial principal and the accumulated interest from previous periods.
The formula used to find the compound interest = [tex]A=P(1+\frac{r}{100})^{nt}[/tex]
Given that, principal=$900, rate of interest=1.5% and time period =11 years.
Now, amount =900(1+1.5/100)¹¹
= 900(1+0.015)¹¹
= 900(1.015)¹¹
= 900×1.177
= $1059.3
Therefore, the amount of money Kadeesha have in her account after 11 years is $1059.3.
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in 1982, the us mint changed the composition of pennies from all copper to zinc with copper coating. pennies made prior to 1982 weigh 3.1 grams. pennies made since 1982 weight 2.5 grams. if you have a bag of 1267 pennies, and the bag weighs 3541.9 grams, how many pennies from each time period are there in the bag?
Prior to 1982, pennies weighed 3.1 grams. Coins produced after 1982 weigh 2.5 grams. 647 coins from each period are there in the bag if it contains 1267 pennies and weighs 3541.9 grams.
Given that,
The 1982 change in pennies composition from all copper to zinc with copper coating was made by the US Mint. Prior to 1982, pennies weighed 3.1 grams. Coins produced after 1982 weigh 2.5 grams.
We have to find how many coins from each period are there in the bag if it contains 1267 pennies and weighs 3541.9 grams.
An issue of quantity and amount. In this context, the term "amount" refers to weight. The variables should be identified with letters that will make them easy to distinguish: Z for largely zinc and C for all copper. Create weight and quantity equations:
C + Z = 1287
3.1 C + 2.5 Z = 3601.5
By multiplying all three components of the first equation by -2.5 and placing them under the second equation, drawing a line, and then subtracting, you can eliminate the Z using this method.
3.1 C + 2.5 Z = 3601.5
-2.5 C - 2.5 Z = -3217.5
--------------------------------
0.6 C = 384
Divide both sides by .6.
C = 640
Subtract 640 from 1287.
Z = 647
Therefore, 647 coins from each period are there in the bag if it contains 1267 pennies and weighs 3541.9 grams.
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Robert has 3 black marbles in a bag, along with 5
marbles of other colours. Robert picks a marble from the bag without looking. What is the probability that Robert picks a black marble?
The probability that Robert picks a black marble is 3/8. The result is obtained from the ratio of the number of black marble to all marbles.
What is probability?Probability is a numerical description of how possible an event to happen. Tha value of probability is between 0 to 1. Zero probability means something is impossible to happen. Probability 1 shows certainty.
Probability of an event can be expressed as
P(A) = n(A) / n(S)
Where
P(A) is the probability of an event An(A) is the number of favorable outcomesn(S) is the total number of events in the sample spaceIn case Robert has 3 black marbles and 5 marbles of other colors, then he picks a marble randomly, what is the probability of picking black marble?
Total number of marbles is 3 + 5 = 8.
n(S) = 8
Number of black marble is 3.
n(A) = 3
The probability of picking black marble is
P(A) = n(A) / n(S)
P(A) = 3/8
Hence, the probability that Robert picks a black marble is 3/8.
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robert john and paul start to run from the same point in the same direction around a circular track.
if they take 126 seconds 154 seconds and 198 seconds respectively to complete one round along the track, when will they next meet again at the starting point?
They will next meet again after 1386 seconds or 23.1 minutes from the starting point.
Given that Robert, John, and Paul begin running in the same direction from the same starting point around a circular track If they take 126 seconds, 154 seconds, and 198 seconds respectively to complete one round of the track.
What is HCF?The highest common factor is defined as the set of numbers with the highest common multiple. The highest positive integer with more than one factor in the set is HCF.
To determine whether they will next meet again at the starting point.
We have to calculate the LCM of the numbers 126, 154, and 198
⇒ Prime factors of 126 = 2 × 2 × 3 ×7
⇒ Prime factors of 154 = 2 × 11 ×7
⇒ Prime factors of 198 = 2 × 3 × 3 ×11
So the LCM of the numbers 126, 154, and 198 will be as:
⇒ 2 × 3 × 3 × 7 × 11 = 1386
Therefore, they will meet after 1386 seconds or 23.1 minutes.
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you are given the set of letters {a, b, c, d, e}. what is the probability that in a random five-letter string (in which each letter appears exactly once, and with all such strings equally likely) the letters a and b are next to each other?
The probability in a random 5-letter string that the letters A and B will be next to each other is 0.4
For A and B to be next to each other in the string of letters, consider the AB combination as a single letter. Then there are 4 letters AB, C, D, E.
There are 4! number of ways to arrange these 4 letters.
4!= 24 ways
We know AB can be arranged in 2 ways ( AB or BA)
Thus there are 2*4! = 2 * 24 = 48 ways of arranging the letters such that AB are next to each other in the random Five-letter string
The total number of letters in the set is 5 and hence the total number of ways the letters can be arranged without A and B necessarily being next to each other is 5!= 120
The probability that in the random five-letter string( in which each letter appears only once with all such strings equally likely) the letters A and B are next to each other= 48/120 = 0.4
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The equation y =1/2x represents a proportional relationship. What is the constant of proportionality? A. 1/2B.XC. 0D. 2
You have the following equation:
y = 1/2 x
In order to determine what is the constant of proportionality, consider the general fom of proportional relatioship, given by:
y = kx
by comparing the previous equation with y=1/2x, you can notice that k=1/2. Then, the constant of proportionality is k = 1/2
A. 1/2
Use inverse trigonometric functions to solve the following equations. If there is more than one solution, enter all solutions as a comma-separated list (like "1, 3"). If an equation has no solutions, enter "DNE".Solve tan(θ)=1 for θ (where 0≤θ<2π).θ=Solve 7tan(θ)=−15 for θ (where 0≤θ<2π).θ=
Starting with the equation:
[tex]\tan (\theta)=1[/tex]take the inverse tangent function to both sides of the equation:
[tex]\begin{gathered} \arctan (\tan (\theta))=\arctan (1) \\ \Rightarrow\theta=\arctan (1) \\ \therefore\theta=\frac{\pi}{4} \end{gathered}[/tex]Yet another value can be found for this equation to be true since the period of the tangent function is π:
[tex]\begin{gathered} \theta_1=\frac{\pi}{4} \\ \theta_2=\frac{\pi}{4}+\pi=\frac{5}{4}\pi \end{gathered}[/tex]Starting with the equation:
[tex]7\tan (\theta)=-15[/tex]Divide both sides by 7:
[tex]\Rightarrow\tan (\theta)=-\frac{15}{7}[/tex]Take the inverse tangent to both sides of the equation:
[tex]\begin{gathered} \Rightarrow\arctan (\tan (\theta))=\arctan (-\frac{15}{7}) \\ \Rightarrow\theta=\arctan (-\frac{15}{7}) \\ \therefore\theta=-1.13416917\ldots \end{gathered}[/tex]The tangent function has a period of π. Since the value that we found for theta is not between 0 and 2π, then we can add π to the value:
[tex]\begin{gathered} \theta_1=-1.13416917\ldots+\pi \\ =2.007423487\ldots \end{gathered}[/tex]We can find another value for theta such that its tangent is equal to -15/7 by adding π again, provided that the result is less than 2π:
[tex]\begin{gathered} \theta_2=\theta_1+\pi \\ =5.14901614\ldots \end{gathered}[/tex]Therefore, for each equation we know that:
[tex]\begin{gathered} \tan (\theta)=1 \\ \Rightarrow\theta=\frac{\pi}{4},\frac{5\pi}{4} \end{gathered}[/tex][tex]\begin{gathered} 7\tan (\theta)=-15 \\ \Rightarrow\theta=2.007423487\ldots\text{ , }5.14901614\ldots \end{gathered}[/tex]Starting with the equation:
[tex]\tan (\theta)=1[/tex]take the inverse tangent function to both sides of the equation:
[tex]\begin{gathered} \arctan (\tan (\theta))=\arctan (1) \\ \Rightarrow\theta=\arctan (1) \\ \therefore\theta=\frac{\pi}{4} \end{gathered}[/tex]Yet another value can be found for this equation to be true since the period of the tangent function is π:
[tex]\begin{gathered} \theta_1=\frac{\pi}{4} \\ \theta_2=\frac{\pi}{4}+\pi=\frac{5}{4}\pi \end{gathered}[/tex]Starting with the equation:
[tex]7\tan (\theta)=-15[/tex]Divide both sides by 7:
[tex]\Rightarrow\tan (\theta)=-\frac{15}{7}[/tex]Take the inverse tangent to both sides of the equation:
[tex]\begin{gathered} \Rightarrow\arctan (\tan (\theta))=\arctan (-\frac{15}{7}) \\ \Rightarrow\theta=\arctan (-\frac{15}{7}) \\ \therefore\theta=-1.13416917\ldots \end{gathered}[/tex]The tangent function has a period of π. Since the value that we found for theta is not between 0 and 2π, then we can add π to the value:
[tex]\begin{gathered} \theta_1=-1.13416917\ldots+\pi \\ =2.007423487\ldots \end{gathered}[/tex]We can find another value for theta such that its tangent is equal to -15/7 by adding π again, provided that the result is less than 2π:
[tex]\begin{gathered} \theta_2=\theta_1+\pi \\ =5.14901614\ldots \end{gathered}[/tex]Therefore, for each equation we know that:
[tex]\begin{gathered} \tan (\theta)=1 \\ \Rightarrow\theta=\frac{\pi}{4},\frac{5\pi}{4} \end{gathered}[/tex][tex]\begin{gathered} 7\tan (\theta)=-15 \\ \Rightarrow\theta=2.007423487\ldots\text{ , }5.14901614\ldots \end{gathered}[/tex]solve the equation, and enter the solutions from least to greatest. If there is only one solution, enter “n.a” for the second solution. (picture of equation listed below)
Answer
x = 1 or x = n.a.
Step-by-step explanation
[tex]\frac{1}{x}+\frac{1}{x-10}=\frac{x-9}{x-10}[/tex]Multiplying by (x - 10) at both sides of the equation:
[tex]\begin{gathered} (x-10)(\frac{1}{x}+\frac{1}{x-10})=\frac{x-9}{x-10}(x-10) \\ \text{ Distributing and simplifying:} \\ \frac{x-10}{x}+\frac{x-10}{x-10}=x-9 \\ \frac{x-10}{x}+1=x-9 \end{gathered}[/tex]Multiplying by x at both sides of the equation:
[tex]\begin{gathered} x(\frac{x-10}{x}+1)=x(x-9) \\ \text{ Distributing and simplifying:} \\ \frac{x(x-10)}{x}+x=x^2-9x \\ x-10+x=x^2-9x \\ 2x-10=x^2-9x \end{gathered}[/tex]Subtracting 2x and adding 10 at both sides of the equation:
[tex]\begin{gathered} 2x-10-2x+10=x^2-9x-2x+10 \\ 0=x^2-11x+10 \end{gathered}[/tex]We can solve this equation with the help of the quadratic formula with the coefficients a = 1, b = -11, and c = 10, as follows:
[tex]\begin{gathered} x_{1,2}=\frac{-b\pm{}\sqrt{b^2-4ac}}{2a} \\ x_{1,2}=\frac{11\pm\sqrt{(-11)^2-4\cdot1\operatorname{\cdot}10}}{2\operatorname{\cdot}1} \\ x_{1,2}=\frac{11\pm\sqrt{81}}{2} \\ x_1=\frac{11+9}{2}=10 \\ x_2=\frac{11-9}{2}=1 \end{gathered}[/tex]The solution x = 10 is not possible because it makes zero the denominator in 2 of the rational expressions of the original equation. In consequence, it must be discarded.
Select whether the pair of lines is parallel, perpendicular, or neither.
y = 5, y = −3
The given pair of lines are parallel lines.
What are parallel lines?Parallel lines in geometry are coplanar, straight lines that don't cross at any point. In the same three-dimensional space, parallel planes are any planes that never cross. Curves with a fixed minimum distance between them and no contact or intersection are said to be parallel. A line with the same slope is said to be parallel. To put it another way, these lines won't ever cross. At the interception, perpendicular lines always intersect and form right angles.So, graph the lines on the given equation:
y = 5y = -3(Refer to the graph attached below)
We can easily look at the graph and tell that the given lines of the equations are parallel to each other.
Therefore, the given pair of lines are parallel lines.
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discuss the advantages and disadvantages of using sampling to reduce the number of data objects. would simple random sampling (without replacement) be a good approach to sampling? why or why not? what kind of sampling method that you would like to use? g
The advantage and disadvantages are sampling.
We have to find the advantages and disadvantages of sampling.
The advantage is:
It provides an opportunity to do data analysis with a lower likelihood of carrying an error.
Researchers can analyze the data that is collected with a smaller margin of error thanks to random sampling. This is permitted since the sampling procedure is governed by predetermined boundaries. The fact that the entire procedure is random ensures that the random sample accurately represents the complete population, which enables the data to offer precise insights into particular topic matters.
The disadvantage is:
No extra information is taken into account.
Although unconscious bias is eliminated by random sampling, deliberate bias remains in the process. Researchers can select areas for random sampling in which they think particular outcomes can be attained to confirm their own bias. The random sampling does not take into account any other information, however sometimes the data collector's supplementary information is retained.
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Please help soon thanks
Answer:
x = -7
Step-by-step explanation:
Hi!
Ok we know a few things here :
x + 37 + x + 67 + 90 = 180
2x + 104 = 90
2x = -14
x = -7
Hope this helps!
Have a great day! :)
Use the following function rule to find g(r + 2). Simplify your answer.
g(k)= k-4
g(r + 2) =
Using the function rule, the value of g(r +2) is r-2
How can we evaluate and simply g(r + 2) using the function rule?Given the function rule g(k) = k-4.
Then g(r + 2) can be found by replacing k with r+2 in the function g(k):
g(r+2) = r + 2 - 4
= r - 2
Therefore, g(r+2) gives r - 2 when evaluated using the function rule.
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The bar graph shows the percentage of college freshmen in a certain country with an average grade of Ain high school.The data displayed by the bar graph can be described by the mathematical model p =4x- + 25 where xis the number of years after 1980 and p is the percentage of college freshmen who had an average
Okay, here we have this:
Considering the provided formula and information, we are going to calculate the requested percent of college freshmen in 2010, so we obtain the following:
So since x corresponds to the number of years after 1980, in this case we will replace x with 30, since it is 30 years after 1980, we have:
p=4x/5+25
Replacing:
p=4(30)/5+25
p=120/5+25
p=24+25
p=49
So, as we can see, the model assumes that in 2010 the percentage was 49%, and the real one is 47%. It means that the model overestimates the real value by 2%.
write the equation of a line through ( -3, 5), parallel to y = -x.
y= ___ x + ___
Answer:
We took it at grade 7 oh god
I'm not good with graphs so I really need help solving and understanding this
4x + 3y = -24 (option D)
Explanation:To determine the orrect option, we would find the equation of the line.
Equation of line: y = mx + c
m = slope, c = y-intercept
Using the slope formula:
[tex]m\text{ = }\frac{y_2-y_1}{x_2-x_1}[/tex]We would use any two points on the graph.
Using points: (-6, 0) and (0, -8)
[tex]\begin{gathered} x_1=-6,y_1=0,x_2=0,y_2\text{ = -}8 \\ m\text{ = }\frac{-8-0}{0-(-6)} \\ m\text{ =}\frac{-8}{0+6}=\frac{-8}{6}=-\frac{4}{3} \end{gathered}[/tex]The y intercept is the point where the line crosses the y axis and the value of x is zero. It crosses the line at y = -8
The equation of line becomes:
y = -4/3 x + (-8)
y = -4/3x - 8
Multiply both sides by 3:
3(y) = 3(-4/3 x) - 3(8)
3y = -4x - 24
3y + 4x = -24
4x + 3y = -24 (option D)
IMAGE ATTACHED,please help me with this one please.
Answer:
Step-by-step explanation:
$17.64 for 147 text messages
Therefore 1 text message will cost $17.64 ÷ 147 Text messages = $0.12 per text message
Find the distance between the points (-16, 20) and (-16, -14).2007(-16, 20)161284-20-16-12-8-4048121620-4-812(-16, -14)-16-20units
Point A
(-16, 20)
Point B
(-16, -14)
The Distance Formula itself is actually derived from the Pythagorean Theorem which is a
[tex]\begin{gathered} d=\sqrt[]{(y2-y1)^2+(x2-x1)^2} \\ d=\sqrt[]{(-14-20)^2+(-16-(-16}))^2 \\ d=\sqrt[]{(-34)^2} \\ d=34 \end{gathered}[/tex]The distance would be 34 units
a candy manufacturer produces bags of jelly beans. the weight of a bag of jelly beans is normally distributed with a mean of 12 ounces and a standard deviation of 0.4 ounces. if a random sample of 4 bags of jelly beans is selected what is the probability that the sample average will be greater than 11.6?
Answer:
2%
Step-by-step explanation:
4x5y² how would I solve this
multiply the numbers
Answer is 20y2
State if the three side lengths form an acute obtuse or a right triangle
Given three side lengths form an acute obtuse or a right triangle 17, 21, & 28
Check Right triangle
[tex]\begin{gathered} Hyp^2=opp^2+adj^2 \\ 28^2=17^2+21^2 \\ 28^2\text{ = 289 +441} \\ 784\text{ }\ne\text{ 730} \end{gathered}[/tex]Not a Right triangle
An obtuse triangle is a triangle with one obtuse angle and two acute angles. Since a triangle's angles must sum to 180° in Euclidean geometry.
Not an Obtuse triangle
[tex]\begin{gathered} \sin \text{ A= }\frac{opp}{hyp} \\ \sin \text{ A = }\frac{17}{28} \\ A=sin^{-1}\text{ }\frac{17}{28} \\ A=37.4^0\text{ (less than 90)} \end{gathered}[/tex][tex]\begin{gathered} \cos \text{ B = }\frac{adj}{hyp} \\ \cos \text{ B = }\frac{21}{28} \\ B=cos^{-1}\frac{21}{28} \\ B=41.4^0\text{ (less than 90)} \end{gathered}[/tex][tex]\begin{gathered} \tan \text{ C = }\frac{opp}{adj} \\ \tan \text{ C = }\frac{17}{21} \\ C=tan^{-1\text{ }}\frac{17}{21}\text{ } \\ C=38.9^0\text{ (less than 90) } \end{gathered}[/tex]Hence it is acute angle because all angles are less than 90°
You are designing a poster with an area of 625 cm2 to contain a printing area in the middle and have the margins of 4cm at the top and bottom and 7cm on each side. Find the largest possible printing area. Round your answer to the nearest four decimal places.
Explanation
Step 1
diagram
so
Step 2
let
a)total area= side1*side 2
[tex]side1*side2=625[/tex]and
[tex]printing\text{ area=\lparen side1-14\rparen\lparen side2-8\rparen}[/tex]Step 3
solve
let
[tex]\begin{gathered} side1\text{ = x} \\ side\text{ 2 =}y \\ xy=625 \\ y=\frac{625}{x}\Rightarrow equation(1) \end{gathered}[/tex]and in teh second equation we have
[tex]\begin{gathered} pr\imaginaryI nt\imaginaryI ng\text{area=}\operatorname{\lparen}\text{s}\imaginaryI\text{de\times1-14}\operatorname{\rparen}\operatorname{\lparen}\text{s}\imaginaryI\text{de\times2-8}\operatorname{\rparen} \\ A=(x-14)(y-8) \\ replace\text{ the y value} \\ A=(x-14)(\frac{625}{x}-8) \\ A=\frac{625x}{x}-\frac{8750}{x}-8x+112) \\ A=-\frac{8750}{x}+625-8x+112 \\ dA=\frac{8750}{x^2}-8 \\ \frac{8750}{x^2}=8 \\ x^2=\frac{8750}{8} \\ x=\sqrt[\placeholder{⬚}]{\frac{8750}{8}} \\ x=33 \end{gathered}[/tex]replace to find y
[tex]\begin{gathered} y=\frac{625}{x}\operatorname{\Rightarrow}equat\imaginaryI on(1) \\ y=\frac{625}{33} \\ y=18.9 \end{gathered}[/tex]so
the maximum area is
[tex]\begin{gathered} x-14=33-14=19 \\ y-8=18.9=10.9 \end{gathered}[/tex]so, the greates area is
[tex]18.9*19=359.1[/tex]therefore, the answer is
359.1 square cm
I hope this helps you
write equations for the horizontal and vertical lines passing through the point (-7,7)
The vertical line's equation is x = -7, and the horizontal line's equation Is y = 7.
Where the given coordinate is (--7, 7).
What do the vertical and horizontal lines represent?A vertical line is a straight, up and down line in a coordinate plane that is parallel to the y-axis. The horizontal line, on the other hand, is parallel to the x-axis and goes straight, left, and right.Here,
The value of x, is -7, will never change for a vertical line. This holds true for any y value. As a result, x = -7 is the equation for a vertical line passing through this point.
Similarly, the value of y, is 7, will never change for a horizontal line. This holds true for any x value. As a result, the equation for a horizontal line passing through this point is as follows: y =7
As a result, the vertical line's equation is x = -7 and the horizontal line's equation is y = 7.
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A magical basket of pennies doubles every day. If the basket if full on the 15th day, on what day was the basket one-quarter full?
The basket of pennies will be 1/4 full on the 13th day
How to determine the day?From the question, the given parameters are:
Rate of change, r = 2 i.e. doubleFull basket, = 15th dayThe above parameters illustrate an exponential function
An exponential function is represented as
A(n) = A * (r)ⁿ⁻¹
Where
A represents the initial value
It is full on the 15th day
So, we have
1 = A *(2)¹⁵⁻¹
Evaluate
1 = A *(2)¹⁴
Divide by (2)¹⁴
A = (2)⁻¹⁴
Substitute A = (2)⁻¹⁴ in A(n) = A * (r)ⁿ⁻¹
A(n) = (2)⁻¹⁴ * (2)ⁿ⁻¹
When it is one-quarter full, we have
1/4 = (2)⁻¹⁴ * (2)ⁿ⁻¹
Multiply by (2)¹⁴
(2)¹² = (2)ⁿ⁻¹
By comparison, we have
n - 1 = 12
Add 1 to both sides
n = 13
Hence, the basket will be one-quarter full on the 13th day
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H=4(x+3y)+2 make x the subject
When the formula is changed to x, the equation is x = 1/4(H - 2) - 3y
How to change the subject of formula to x?From the question, the equation is given as
H=4(x+3y)+2
Subtract 2 from both sides of the equation
So, we have the following equation
H - 2 =4(x+3y)+2 -2
Evaluate
So, we have the following equation
H - 2 =4(x+3y)
Divide by 4
So, we have the following equation
1/4(H - 2) = x +3y
Subtract 3y from both sides
x = 1/4(H - 2) - 3y
Hence, the solution for x is x = 1/4(H - 2) - 3y
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The simplified solution for the x is given as x = [H - 2]/4 - 3y.
As mentioned in the question, for an equation H=4(x+3y)+2, we have to transform the equation in the subject of x.
The operational processes in mathematics for the functions to make the function or expression.
Here,
Simplification of the given function in a manner to the transposition of the variables from left to right,
H=4(x+3y)+2
H - 2 = 4(x+3y)
{H - 2}/4 = (x + 3y)
x = [H - 2]/4 - 3y
Thus, the simplified solution for the x is given as x = [H - 2]/4 - 3y.
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Find where the graphs intersect; f(x)=2x+3 and g(x)=-0.5x+7
Step 1
Given
[tex]\begin{gathered} f(x)=\text{ 2x+3} \\ g(x)\text{ = -0.5x+7} \end{gathered}[/tex]Required: To find where the graph of both functions intersect. In other words the find the value of x and hence f(x) and g(x).
Step 2
Solve both equations simultaneously.
[tex]\begin{gathered} we\text{ will take f(x) and g(x) = y, so that} \\ y=2x+3\text{ -----(1)} \\ y=-0.5x+7----(2) \\ \end{gathered}[/tex]Subtract equation 2 from 1
Hence,
[tex]\begin{gathered} 4=\text{ 2.5x} \\ \frac{4}{2.5}=\frac{2.5x}{2.5} \\ x\text{ = 1.6} \end{gathered}[/tex]Step 3
Check
[tex]\begin{gathered} f(x)\text{ = 2x+3} \\ f(1.6)=\text{ 2(1.6) + 3 = 6.2} \\ g(x)=\text{ -0.5x+7} \\ g(1.6)=\text{ -0.5(1.6) +7 = 6.2} \\ \text{since the check gave us the same values, x = 1.6} \\ \text{And the coordinate point of the solution will be ( 1.6, 6.2)} \end{gathered}[/tex]Hence the graph intersects at the point where x = 1.6 and y =6.2. Remember y = f(x) and g(x)
Number 6. For questions 5-7, (a) use synthetic division to show that x is a zero.(b) find the remaining factors of f(x).(c) use your results to find the complete factorization of f(x).(d) list all zeros of f(x).(e) graph the function.
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given polynomials
[tex]f(x)=x^3+6x^2-15x-100[/tex]One of the zeroes is:
[tex]\begin{gathered} x=-5 \\ \text{this implies that:} \\ (x+5)=0 \end{gathered}[/tex]STEP 2: Use synthetic division to divide the polynomials
[tex]\frac{x^3+6x^2-15x-100}{x+5}[/tex]Write the coefficients of the numerator
[tex]1\:\:6\:\:-15\:\:-100[/tex][tex]\begin{gathered} \mathrm{Write\:the\:problem\:in\:synthetic\:division\:format} \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \\ Carry\:down\:the\:leading\:coefficient,\:unchanged,\:to\:below\:the\:division\: \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \\ \end{gathered}[/tex][tex]\begin{gathered} Multiply\:the\:carry-down\:value\:by\:the\:zero\:of\:the\:denominator,\:and\:carry\:the\:result\:up\:into\:the\:next\:column \\ 1\left(-5\right)=-5 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{Add\:down\:the\:column:} \\ 6-5=1 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:\:\:\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} Multiply\:the\:carry-down\:value\:by\:the\:zero\:of\:the\:denominator,\:and\:carry\:the\:result\:up\:into\:the\:next\:column: \\ 1\left(-5\right)=-5 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{Add\:down\:the\:column:} \\ -15-5=-20 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:\:\:\:}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:-20\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} Multiply\:the\:carry-down\:value\:by\:the\:zero\:of\:the\:denominator,\:and\:carry\:the\:result\:up\:into\:the\:next\:column: \\ \left(-20\right)\left(-5\right)=100 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:100}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:-20\:\:\:\:\:\:}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{Add\:down\:the\:column:} \\ -100+100=0 \\ \begin{matrix}\texttt{\:\:\:\:-5¦\:\:\:\:\:1\:\:\:\:\:6\:\:\:-15\:\:-100}\\ \texttt{\:\:\:\:\:\:¦\underline{\:\:\:\:\:\:\:\:\:\:-5\:\:\:\:-5\:\:\:100}}\\ \texttt{\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:1\:\:\:-20\:\:\:\:\:0}\end{matrix} \end{gathered}[/tex][tex]\begin{gathered} \mathrm{The\:last\:carry-down\:value\:is\:the\:remainder} \\ 0 \end{gathered}[/tex]The last carry-down value is the remainder and it is 0 (zero)
Since the remainder is a zero, hence, x=-5 is a zero
Step 3: Answer question b
To get the factors, the remainder of the division in step 2 is given as:
The remaining factors of f(x) is:
[tex]x^2+x-20[/tex]STEP 4: Answer Question c
[tex]\begin{gathered} roots=(x+5)(x^2+x-20) \\ Factorize\text{ the other root to have:} \\ Using\text{ factorization methods:} \\ (x^2+x-20)=(x^2+5x-4x-20) \\ x(x+5)-4(x+5)=0 \\ (x-4)(x+5)=0 \end{gathered}[/tex]The complete factorization will give:
[tex](x+5)(x-4)(x+5)[/tex]STEP 5: Answer question d
The zeroes of f(x) will be:
[tex]\begin{gathered} zeroes\text{ of f\lparen x\rparen=?, we equate the roots to 0} \\ zeroes\Rightarrow x=-5,4,-5 \end{gathered}[/tex]zeroes are: -5,4,-5
STEP 6: Plot the graph