The GCF of these monomials i.e, 28g^5h^2 and 12g^6h^5 is 4h^2g^5
What is monomials?
Monomial expressions include only one non-zero term. Numbers, variables, or multiples of numbers and variables are all examples of monomials.
First take the coefficient ie, 28 and 12 to find the GCF
The GCF of 28 and 12 is 4
Now, find out the GCF of the variables for that you take the lowest exponent from both the variables g and h
for g variable it will be g^5 and,
for h variable it will be h^2
Therefore, the GCF of these monomials is 4h^2g^5
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How many roots does x^2-6x+9 have ? It may help to graph the equation.
The roots are those values that make a function or polynomial take a zero value. The roots are also the intersection points with the x-axis. In the case of a quadratic equation you can use the quadratic formula to find its roots:
[tex]\begin{gathered} ax^2+bx+c=y\Rightarrow\text{ Quadratic equation in standard form} \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}\Rightarrow\text{ Quadratic formula} \end{gathered}[/tex]So, in this case, you have
[tex]\begin{gathered} y=x^2-6x+9 \\ a=1 \\ b=-6 \\ c=9 \end{gathered}[/tex][tex]\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(1)(9)}}{2(1)} \\ x=\frac{6\pm\sqrt[]{36-36}}{2} \\ x=\frac{6\pm0}{2} \\ x=\frac{6}{2} \\ x=3 \end{gathered}[/tex]As you can see, this function only has one root, at x = 3.
You can see this in the graph of the function:
PLEASE DO IT ASAP
What is the value of the expression?
0.3(1/4 - 1) + 0.35
-0.575
-0.125
0.125
1.4
1.925
The value of the expression 0.3(1/4 - 1) + 0.35 is 0.125
The expression is
0.3(1/4 - 1) + 0.35
The expression is defined as the sentence with a minimum of two variables and at least one math operation.
Here the expression is
0.3 (1/4 - 1) + 0.35
First do the arithmetic operation in the bracket
0.3(1/4 - 1) + 0.35 = 0.3 × -0.75 + 0.35
In next step do the multiplication
0.3 × -0.75 + 0.35 = -0.225 + 0.35
Do the addition of the numbers
-0.225 + 0.35 = 0.125
Hence, the value of the expression 0.3(1/4 - 1) + 0.35 is 0.125
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1) There is a proportional relationship between the number of months a person has had a streaming movie subscription and the total amount of money they have paid for the subscription. The cost for 6 months is $47.94. The point (6,47.94) is shown on the graph below. 180 160 140 120 100 cost (dollars) 80 60 (6, 47.94) 40 20 16 18 8 20 22 2. 4 6 10 12 14 time (months)
Given:
The point which describes the relationship between the months and total amount is, (6, 47.94).
a) To find the constant proportionality:
6 months =47.94
Then, for 1 month,
[tex]\frac{47.94}{6}=7.99[/tex]Hence, the constant proportionality is $7.99.
b) The constant proportionality tells that, if the month is increased then the cost is also increased by $7.99.
c) To find the three more points and label it:
For the month, m=1, then the cost c=$7.99
For the month, m=2, then the cost c=$15.98
For the month m=3, then the cost c=$23.97
Therefore, the three points are (1, 7.99), (2,15.98) and (3, 23.97).
The graph is,
d) The relationship between the months and the cost is,
C=7.99 m
3, -10, 16, -36, 68, ___-3, 12, -33, 102, -303, ___Identify a pattern in each list of numbers. Then use this pattern to find the next number.
As for the sequence 3,-10,16,-36,68,..., notice that
[tex]\begin{gathered} 3-13=-10 \\ -10+26=-10+2(13)=-10+2^1(13)=16 \\ 16-52=16-4(13)=16-2^2(13)=-36 \\ -36+104=-36+8(13)=-36+2^3(13)=68 \end{gathered}[/tex]Therefore, the next term is
[tex]68-2^4(13)=68-16(13)=-140[/tex]The answer is -140.
Regarding the second pattern, notice that
[tex]\begin{gathered} -3+15=12 \\ 12-45=12-3(15)=12-3^1(15)=-33 \\ -33+135=-33+9(15)=-33+3^2(15)=102 \\ 102-405=102-27(15)=102-3^3(15)=-303 \end{gathered}[/tex]Then, the next term of the sequence is
[tex]-303+3^4(15)=912[/tex]The answer is 912
A student entering a doctoral program in educational psychology is required to select two courses from the list provided as part of his or her program (a)List all possible two-course selections (b)Comment on the likelihood that you EPR 625 and EPR 686 will be selected The course list EPR 613, EPR 664, EPR 625, EPR 685, EPR 686(a)select all the possible two-course selections belowA. 613, 686B. 625,686C. 613,613,664D. 664,685E. 625,685F. 625,672G. 613,625H. 685,686I. 664,625J 686,686K. 613,613L. 613,685M. 664, 686N. 613,664
List of courses that the student entering a doctoral program in educational psychology can take:
EPR 613, EPR 664, EPR 625, EPR 685, EPR 686
Therefore, the possible two-course selections for the student are:
A. Both courses are on the list given: 613, 686
B. Both courses are on the list given: 625, 686
C. It's not possible. This option contains three courses.
D. Both courses are on the list given: 664, 685
E. Both courses are on the list given: 625, 685
F. It's not possible, Course 672 isn't available.
G. Both courses are on the list given: 613, 625
H. Both courses are on the list given: 685, 686
I. Both courses are on the list given: 664, 625
J. It's not possible. Just one course is given.
K. Same case than J. Just one course.
L. Both courses are on the list given: 613, 685
M. Both courses are on the list given: 664, 686
N. Both courses are on the list given: 613, 664
#17 - A bin contains 90 batteries (all size C). There are 30 Eveready, 24 Duracell, 20 Sony,10 Panasonic, and 6 Rayovac batteries. What is the probability that the battery selected is aDuracell?0 27.6%0 26.7%24.6%0 29.2%
According to the basic definition of probability,
[tex]\text{Probability}=\frac{\text{ No. of favorable events}}{\text{ Total no. of events}}[/tex]Given that the bin contains total 90 batteries, out of which 24 are duracell.
So the probability that a randomly selected battery is Duracell, is calculated as,
[tex]\begin{gathered} P(\text{Duracell)}=\frac{\text{ No. of Duracell Batteries}}{\text{ Total no. of batteries}} \\ P(\text{Duracell)}=\frac{24}{90} \\ P(\text{Duracell)}\approx0.267 \\ P(\text{Duracell)}\approx26.7\text{ percent} \end{gathered}[/tex]Thus, the probability that a randomly selected battery is Duracell, is 26.7% approximately.
Given a triangle ABC at points A = ( - 2, 2 ) B = ( 2, 5 ) C = ( 2, 0 ), and a first transformation of right 4 and up 3, and a second transformation of left 2 and down 5, what would be the location of the final point B'' ?
Answer
a. (4, 3)
Step-by-step explanation
The translation of a point (x, y) a units to the right and b units up transforms the point into (x + a, y + b).
Considering point B(2, 5), translating it 4 units to the right and 3 units up, we get:
B(2, 5) → (2+4, 5+3) → B'(6, 8)
The translation of a point (x, y) c units to the left and d units down transforms the point into (x - c, y - d).
Considering point B'(6, 8), translating it 2 units to the left and 5 units down, we get:
B'(6, 8) → (6 - 2, 8 - 5) → B''(4, 3)
Answer: The answer would be (4,3)
Step-by-step explanation: because if you started with (2,5), which would be (x,y) x goes left and right, and y goes up and down, and the questions says that you have to go 4 to the right and 3 up, then add 4 to 2, which is 6, and 3 to 5, which is 8, so now you have the point (6,8), then the second translation would be 2 to the left, and down 5, this is negative so you subtract this time, so subtract 2 from 6, which is 4, and 5 from 8, which is 3, so your final answer is (4,3).
The total fixed costs of producing a product is $55,000 and the variable cost is $190 per item. If the company believes they can sell 2,500 items at $245 each, what is thebreak-even point?800 items900 items960 items 1,000 itemsNone of these choices are correct.
Let's call FC the fixed cost for production and VC the variable cost per item.
The company believes they can sell 2,500 items at $245 each.
Production costs:
For producing 2,500 items, the company has to spend (total cost, TC):
[tex]\begin{gathered} TC=FC+2,500\cdot VC \\ TC=55,000+2,500\cdot190 \\ TC=530,000 \end{gathered}[/tex]Sells:
Now, company sells eacho of the 2,500 items at $245, so, the company income (I) is:
[tex]I=245\cdot x[/tex]where x is the number of items sold.
Break-even point:
This point is reached when company can recover the money they spend (TC). So, we have the following eaquation to solve:
[tex]\begin{gathered} TC\text{ = I} \\ \to530,000=245\cdot x \\ \to x=\frac{530,000}{245}\text{ =2,163.3 (rounded) } \end{gathered}[/tex]Since company can not sell fractions of items, they have to sell 2,164 items to take back the money they invested.
So, "None of these choices are correct".
Select the sequence of transformations that will carry rectangle A onto rectangle A'. A) reflect over y-axis, rotate 90° clockwise, then reflect over x-axis B) rotate 180° clockwise, reflect over y-axis, then translate 3 units left C) rotate 180° clockwise, reflect over x-axis, then translate 2 units left D) rotate 90° clockwise, reflect over y axis, then translate 3 units left
Let:
[tex]\begin{gathered} A=(3,4) \\ B=(4,2) \\ C=(1,-1) \end{gathered}[/tex]and:
[tex]\begin{gathered} A^{\prime}=(-3,1) \\ B^{\prime}=(-4,-1) \\ C^{\prime}=(-1,-4) \end{gathered}[/tex]After a reflection over the y-axis:
[tex]\begin{gathered} A\to(-x,y)\to A_1=(-3,4) \\ B\to(-x,y)\to B_1=(-4,2) \\ C\to(-x,y)\to C_1=(-1,-1) \end{gathered}[/tex]After a translation 3 units down:
[tex]\begin{gathered} A_1\to(x,y-3)\to A_2=(-3,1) \\ B_1\to(x,y-3)\to B_2=(-4,-1) \\ C_1\to(x,y-3)\to C_2=(-1,-4) \end{gathered}[/tex]Since:
[tex]\begin{gathered} A_2=A^{\prime} \\ B_2=B^{\prime} \\ C_2=C^{\prime} \end{gathered}[/tex]The answer is the option K.
A model of a dinosaur skeleton was made using a scale of 1 in : 15 in in a museum. If the size of the dinosaur’s tail in the model is 8 in, then find the actual length of dinosaur’s tail.
The length of the real dinosaur's tail is 120 inches.
How to find the actual length of the tail?We know that the scale of the model is 1in : 15in, this means that each inch in the model represents 15 inches of the actual dinosaur.
So, if the tail of the model has a length of 8 inches, the length of the real tail will have 15 times that, so the length is given by the product:
8in*15 = 120in
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Question 3 10 pts When solving an absolute value equation, such as |2x + 51 = 13, it is important to create two equations: 2x + 5= [ Select] and 2.1 + 5 = [Select ] [ Select] Resulting in z = vor [Select] Question 4 5 pts
1) Solving that absolute value equation:
|2x+5|=13 Applying the absolute value eq. property
2x +5 = 13 subtracting 5 from both sides
2x = 13-5
2x= 8 Dividing by 2
x =4
2x +5=-13 subtracting 5 from both sides
2x = -13-5
2x = -18 Dividing by 2
x= -9
Then x=4 or x =-9
2) The equations 2x +5 =13 and 2x +15= -13
Resulting in x=4 or x =-9
Production has indicated that they can produce widgets at a cost of $16.00 each if they lease new equipment at a cost of $40,000. Marketing has estimated the number of units they can sell at a number of prices (shown below). Which price/volume option will allow the firm to avoid losing money on this project?
The price/volume option that will allow the firm to avoid losing money on this project is C. 2,300 units at $34.00 each.
How is this option determined?To determine the correct option, we use the cost-volume-profit analysis tool.
The cost-volume-profit (CVP) analysis involves determining how the volume of sales drives profitability.
The CVP technique classifies costs into their variable and fixed cost elements for the purpose of this analysis.
Variable cost per unit = $16
Fixed cost = $40,000
Option A Option B Option C Option D Option E
Sales units 3,000 1,900 2,300 2,500 1,700
Unit selling price $29 $36.50 $34 $31.50 $39
Sales revenue $87,000 $69,350 $78,200 $78,750 $66,300
Variable costs 48,000 30,400 36,800 40,000 27,200
Fixed cost 40,000 40,000 40,000 40,000 40,000
Total costs 88,000 70,400 76,800 80,000 67,200
Thus, the price/volume option that meets the firm's goal is Option C because the sales revenue exceeds the total costs.
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Question Completion with Price/Volume Options:A. 3,000 units at $29.00 each.
B. 1,900 units at $36.50 each.
C. 2,300 units at $34.00 each.
D. 2,500 units at $31.50 each.
E. 1,700 units at $39.00 each.
I need help if u need a pic of the graph I’ll take a picture of it
A.
Using the points (2,3) and (0,6) to find the slope (m), we have:
[tex]m=\frac{y2-y1}{x2-x1}=\frac{6-3}{0-2}=\frac{3}{-2}[/tex]The slope is m= -3/2
B.
Using the points (-1, 7.5) and (1, 4.5) to find the slope (m), we have:
[tex]m=\frac{y2-y1}{x2-x1}=\frac{4.5-7.5}{1-(-1)}=\frac{-3}{1+1}=\frac{-3}{2}[/tex]The slope is m= -3/2
C.
The slope is the same as we are finding the ratio of the vertical change to the horizontal change between two points. Since the function represents a linear equation the slope is going to be the same despite of the points you choose.
Number 5 need help I really forgot how to solve this problem
Line Segments and Rays
A line segment has two endpoints. It contains these endpoints and all the points of the line between them,
A ray is a part of a line that has one endpoint and goes on infinitely in only one direction. You cannot measure the length of a ray.
The figure shows a line that starts in B and goes infinitely to the left side, passing through A, thus the correct choice is B. Ray BA
Abdul will rent a car for a day. The rental company offers two pricing options: Option A and Option B. For each pricing option, cost (in dollars) depends on miles driven, as shown below.
From the graph, we are to determine the following:
(a) We are to find the option that costs less if Abdul drives 300 miles of the rental car and also how much less is it from the other option.
Option A: when x = 300, y = 140
Option B: when x = 300, y = 120
So the difference is:
140 - 120 = 20
So the option that costs less is B
And it costs $20 lesser than option A
(b) For what number of miles does the option costs the same and if Abdul drives less than that amount, what option cost more.
Option A: when x = 100, y = 60
Option B: when x = 100, y = 60
Therefore, the number of miles where the options cost the same is 100 miles.
If Abdul drives less than the amount:
That is x < 100, the B > A,
Which means, if Abdul drives less than 100 miles, Option B, costs more.
a teacher bought 4 folders and 9 books for $33.75. on another day, she bought 3 folders and 12 books at the same prices for $34.50. how much did she pay for each folder and each book?
The teacher made two different purchases:
First purchase:
4 folders and 9 books for $33.75
Second purchase
3 folders and 12 books for $34.50
Let "f" represent the cost of each folder and "b" represent the cost of each book. You can express the total cost of each purchase as equations:
[tex]\begin{gathered} 1)4f+9b=33.75 \\ 2)3f+12b=34.50 \end{gathered}[/tex]Now we have established a system of equations, to solve it, the first step is to write one of the equations in terms of one of the variables.
For example, I will write the first equation in terms if "f"
[tex]\begin{gathered} 4f+9b=33.75 \\ 4f=33.75-9b \\ \frac{4f}{4}=\frac{33.75-9b}{4} \\ f=\frac{135}{16}-\frac{9}{4}b \end{gathered}[/tex]The second step is to replace the expression obtained for "f" in the second equation:
[tex]\begin{gathered} 3f+12b=34.50 \\ 3(\frac{135}{16}-\frac{9}{4}b)+12b=34.50 \end{gathered}[/tex]Distribute the multiplication on the parentheses term
[tex]\begin{gathered} 3\cdot\frac{135}{16}-3\cdot\frac{9}{4}b+12b=34.50 \\ \frac{405}{16}-\frac{27}{4}b+12b=34.50 \\ \frac{405}{16}+\frac{21}{4}b=34.50 \end{gathered}[/tex]Pass the number to the right side of the equal sign by applying the opposite operation to both sides of it
[tex]\begin{gathered} \frac{405}{16}-\frac{405}{16}+\frac{21}{4}b=34.50-\frac{405}{16} \\ \frac{21}{4}b=\frac{147}{16} \end{gathered}[/tex]Now divide b by 21/4 to cancel the multiplication and to keep the equality valid, you have to divide both sides of the expression, so divide 147/16 by 21/4 too, or multiply them by its reciprocal fraction, 4/21, is the same.
[tex]\begin{gathered} (\frac{21}{4}\cdot\frac{4}{21})b=(\frac{4}{21}\cdot\frac{147}{16}) \\ b=\frac{7}{4}\approx1.75 \end{gathered}[/tex]Each book costs $1.75
Now that we have determined how much does each book cost, we can determine the cost of each folder by replacing the value of "b" in the expression obtained for "f"
[tex]\begin{gathered} f=\frac{135}{16}-\frac{9}{4}b \\ f=\frac{135}{16}-\frac{9}{4}\cdot\frac{7}{4} \\ f=\frac{9}{2}\approx4.5 \end{gathered}[/tex]Each folder costs $4.50
Function g can be thought of as a translated (shifted) version of f(x) = x?Y Y6+5+432f7 6 5 4 3 21 2 3 4 5 6 7-2--3+-6-7Write the equation for g(x).
Answer:
g(x) = (x + 5)²
Explanation:
g is the same function f shifted 5 units to the left.
Then, if we have a function h(x) =f(x+c), h(x) is f(x) shifted c units to the left.
So, to translate f 5 units to the left, we need to replace x by (x + 5), to get:
[tex]\begin{gathered} f(x)=x^2 \\ g(x)=f(x+5) \\ g(x)=(x+5)^2 \end{gathered}[/tex]So, the equation for g(x) is:
g(x) = (x + 5)²
A forest products company claims that the amount of usable lumber in its harvested trees averages142 cubic feet and has a standard deviation of 9 cubic feet. Assume that these amounts haveapproximately a normal distribution.1. What percent of the trees contain between 133 and 169 cubic feet of lumber? Round to twodecimal places.II. If 18,000 trees are usable, how many trees yield more than 151 cubic feet of lumber?
1) Considering that the amount of lumber in this Data Set has been normally distributed, then we can start by finding this Percentage (or probability in this interval, writing out the following expressions:
[tex]\begin{gathered} P(133Now we can replace it with the Z score formula, plugging into that the Mean, the Standard Deviation, and the given values:[tex]Z=\frac{X-\mu}{\sigma}[/tex]Then:
[tex]\begin{gathered} P(\frac{133-142}{9}<\frac{X-\mu}{\sigma}<\frac{169-142}{9}) \\ P(-1Checking a Z-score table we can state that the Percentage of the trees between 133 and 169 ft³ is:[tex]P(-12) Now, let's check for the second part, the number of trees. But before that, let's use the same process to get a percentage that fits into that:[tex]\begin{gathered} P(X>151)=\frac{151-142}{9}=1 \\ P(Z>1)=0.1587 \end{gathered}[/tex]Note that 0.1587 is the same as 15.87%. Multiplying that by the total number of trees we have:
[tex]18000\times0.1587=2856.6\approx2857[/tex]Rounding it off to the nearest whole.
3) Thus, The answers are:
i.84%
ii. 2857 trees
A house casts a shadow that is 12 feet tall. A woman who is 5.5 feet tall casts a shadow that is 3 feet tall.
What is the height of the house?
A. 22 ft.
B. 55 ft.
C. 5.5 ft.
D.220 ft.
jessica bought 4 gallons of paint. Jessica needed to use 3/4 of the paint to paint her living room and dining room. How many gallons did she use, write the number of gallons.
Jessica bought 4 gallons of paint. Of that, she used 3/4 to paint. So the ammount she used was
[tex]4\cdot(\frac{3}{4})=\frac{4\cdot3}{4}=3[/tex]So she used 3 gallons of paint.
Read the problem below and find the solution. Use a model or act the
problem out to help solve it.
A group of 24 students have recess together. They are making teams to play
a game. Each team has to have exactly 5 players, and no one can be on more
than one team. How many teams can they make? (It is possible that not
everyone can be on a team.)
Answer:
possible
Step-by-step explanation:
3. You draw one card from a standard deck.(a) What is the probability of selecting a king or a queen? (b) What is the probability of selecting a face card or a 10? (c) What is the probability of selecting a spade or a heart? (d) What is the probability of selecting a red card or a black card?
Given:
The objective is to find,
a) The probability of selecting a king or a queen.
b) The probability of selecting a face card or a 10.
Explanation:
The total number of cards in a deck is, N = 52 cards.
a)
Out of 52 cards, the number of king cards is,
[tex]n(k)=4[/tex]Similarly, out of 52 cards, the number of queen cards is,
[tex]n(q)=4[/tex]Then, the probability of drawing one out of 4 king cards or one out of 4 queen cards can be calculated as,
[tex]\begin{gathered} P(E)=P(k)+P(q) \\ =\frac{n(k)}{N}+\frac{n(q)}{N} \\ =\frac{4}{52}+\frac{4}{52} \\ =\frac{8}{52} \end{gathered}[/tex]Hence, the probsability of selecting a king or a queen is (8/52).
b)
Out of 52 cards, the number of face cards is 12.
[tex]n(f)=12[/tex]Similarly, out of 52 cards, the number of 10 is,
[tex]n(10)=4[/tex]Then, the probability of drawing one out of 12 face cards or one out of 4 ten cards can be calculated as,
[tex]\begin{gathered} P(E)=P(f)+P(10) \\ =\frac{12}{52}+\frac{4}{52} \\ =\frac{12+4}{52} \\ =\frac{16}{52} \end{gathered}[/tex]Hence, the probability of selecting a face card or a 10 is (16/52).
c)
Out of 52 cards, the number of spade cards is 13.
[tex]n(s)=13[/tex]Similarly, out of 52 cards, the number of heart cards is 13.
[tex]n(h)=13[/tex]Then, the probability of drawing one out of 13 spade cards or one out of 13 heart cards can be calculated as,
[tex]\begin{gathered} P(E)=P(s)+P(h) \\ =\frac{n(s)}{N}+\frac{n(h)}{N} \\ =\frac{13}{52}+\frac{13}{52} \\ =\frac{26}{52} \end{gathered}[/tex]Hence, the probability of selecting a spade or a heart is 26/52.
d)
Out of 52 cards, the number of red cards is,
[tex]n(r)=26[/tex]Out of 52 cards, the number of black cards is,
[tex]n(b)=26[/tex]Then, the probability of drawing one out of 26 red cards or one out of 26 black cards is,
[tex]\begin{gathered} P(E)=P(r)+P(b) \\ =\frac{n(r)}{N}+\frac{n(b)}{N} \\ =\frac{26}{52}+\frac{26}{52} \\ =\frac{52}{52} \\ =1 \end{gathered}[/tex]Hence, the probability of selecting a red card or a black card is 1.
Solve. 4 + x/7 = 2Question 3 options:12-144210
1) Since we have a Rational Equation let's proceed with that, isolating the x on one side and then we can get rid of that fraction. This way:
[tex]\begin{gathered} 4+\frac{x}{7}=2 \\ 4-4+\frac{x}{7}=2-4 \\ \frac{x}{7}=-2 \end{gathered}[/tex]Notice that now, we're going to get rid of that fraction on the left side, multiplying it by 7 (both sides) :
[tex]\begin{gathered} 7\times\frac{x}{7}=-2\times7 \\ x=-14 \end{gathered}[/tex]Thus, the answer is -14
Let f(t) = 3 + 2, g(x) = -x^2?, andhe) = (x - 2)/5. Find the indicated value:24. h (g(5))
The Solution to Question 24:
Given the function below:
[tex]\begin{gathered} g(x)=-x^2 \\ h(x)=\frac{x-2}{5} \end{gathered}[/tex]We are asked to find the value of h(g(5)).
Step 1:
We shall find g(5) by substituting 5 for x in g(x).
[tex]g(5)=-5^2=-25[/tex]So that:
[tex]h(g(5))=h(-25)[/tex]Similarly, we shall find h(-25) by substituting -25 for x in h(x).
[tex]h(-25)=\frac{-25-2}{5}=\frac{-27}{5}[/tex]Therefore, the correct answer is
[tex]\frac{-27}{5}[/tex]For the function f(x)=3x2−4x−4,a. Calculate the discriminant.b. Determine whether there are 0, 1, or 2 real solutions to f(x)=0.
Answer:
a) Using the formula for the discriminant we get:
[tex]\begin{gathered} \Delta=(-4)^2-4(3)(-4), \\ \Delta=16+48, \\ \Delta=64. \end{gathered}[/tex]The discriminant is 64.
b) Based on the above result we know that the f(x)=0 has 2 real solutions,
solve p(x+q)^4=r for x
Given the following equation:
[tex]p\mleft(x+q\mright)^4=r[/tex]You can solve for the variable "x" as following:
1. You need to apply the Division property of equality by dividing both sides of the equation by "p":
[tex]\begin{gathered} \frac{p\mleft(x+q\mright)^4}{p}=\frac{r}{p} \\ \\ \mleft(x+q\mright)^4=\frac{r}{p} \end{gathered}[/tex]2. Remember that:
[tex]\sqrt[n]{a^n}=a[/tex]Then:
[tex]\begin{gathered} \sqrt[4]{(x+q)^4}=\sqrt[4]{\frac{r}{p}} \\ \\ x+q=\sqrt[4]{\frac{r}{p}} \end{gathered}[/tex]3. Now you have to apply the Subtraction property of equality by subtracting "q" from both sides of the equation:
[tex]\begin{gathered} x+q-(q)=\sqrt[4]{\frac{r}{p}}-(q) \\ \\ x=\sqrt[4]{\frac{r}{p}}-q \end{gathered}[/tex]The answer is:
[tex]x=\sqrt[4]{\frac{r}{p}}-q[/tex]The safe load, L, of a wooden beam of width w, height h, and length l, supported at both ends, varies directly as the product of the width and the square of the height, and inversely as the length. A wooden beam 5 inches wide, 8 inches high, and 216 inches long can hold a load of 7670 pounds. What load would a beam 3 inches wide, 5 inches high, and 240 inches long of the same material, support? Round your answer to the nearest integer if necessary.
we know that
L=KW(h^2)/l
we have that
W=5 in
h=8 in
l=216 in
L=7670 pounds
step 1
Find the value of K (constant of proportionality)
substitute the given values in the equation
7670=K(5)(8^2)/216
7670=k(1.4815)
k=5,177.25
step 2
we have the equation
L=(5,177.25)W(h^2)/l
for
W=3 in
h=5 in
l=240 in
substitute in the equation and solve for L
L=(5,177.25)(3)(5^2)/240
L=1,617.89 pounds
Round your answer to the nearest integer
so
L=1,618 pounds
there are 150 oranges in 10 craftes of each crate has the same amount of oranges how many oranges are in each crate?
Take the total number of oranges and divide by the number of crates
150 orange
----------------
10 crates
15 oranges per crate
Ary is writing thank you cards to everyone who came to her wedding. It takes her of an hour to write one thank you card. If it took her 8 hours to finish writing all of the cards, how many thank you cards did she write? 48 thank you cards 36 thank you cards 46 thank you cards 40 thank you cards
The question doesn't specify which fraction of an hour it takes Ary to write a thank you card.
Let's imagine that it takes her 1/4 of an hour to write a thank you card.
In such case, in one hour she will be able to write 4 thank you cards.
and therefore in 8 hours, ishe will be able to write 32 thank you cards (8 times 4 cards).
If it takes her 1/6 of an hour to write a thank you card, then in hone hour she will write a total of 6 thank you cards, and therefore, in 8 hours she will be able to write 8 times 6 thank you cards: 8 x 6 = 48 thank you cards.
If it takes her 1/5 of an hour to write a thank you card, then in hone hour she will write a total of 5 thank you cards, and therefore, in 8 hours she will be able to write 8 times 5 thank you cards: 8 x 5 = 40 thank you cards.
You just use this type of criteria to solve the problem whatever the fraction of the hour it takes to write one card as they specify in the question.
Let's find2. 1+5 3First write the addition with a common denominator.Then add.12— +51-4-13Х5
The given addition exercise is:
[tex]\frac{2}{5}+\frac{1}{3}[/tex]The LCM of the denominator (5 and 3) = 15
Multiply 2/5 by 3/3
[tex]\frac{2}{5}=\frac{2\times3}{5\times3}=\frac{6}{15}[/tex]Multiply 1/3 by 5/5
[tex]\frac{1}{3}=\frac{1\times5}{3\times5}=\frac{5}{15}[/tex]The addition becomes
[tex]\frac{6}{15}+\frac{5}{15}=\frac{11}{15}[/tex]Therefore, we can fill in the vacant boxes as shown below:
[tex]\frac{2}{5}+\frac{1}{3}=\frac{6}{15}+\frac{5}{15}=\frac{11}{15}[/tex]