RATIONAL FUNCTIONSSynthetic divisiontable buand write your answer in the following form: Quotient *
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The given polynomial is:
[tex]\frac{2x^4+4x^3-6x^2+3x+8}{x\text{ + 3}}[/tex]Using the long division method:
The equattion can be written in the form:
Quotient + Remainder / Divisor
[tex](2x^3-2x^2\text{ + 3) +}\frac{-1}{x+3}[/tex]
Related Questions
H is the circumcenter of triangle BCD, BC=18, and HD=14. Find CH.
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Given that H is the circumcenter of the triangle.
It means, the length between each vertex point of the triangle and the point H is the radius of the circle.
Thus, the line DH=CH=BH are the radius of the circle.
It is given that DH=14.
Therefore CH=14.
Hence the value of CH is 14.
I didn't really get it when my teacher tried to explain this
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The formula for determining the volume of a cylinder is expressed as
V = pi * r^2h
Where
V represents volume of cylinder
pi is a constant whose value is 3.142
r represents radius of cylinder
h represents height of cylinder
From the information given,
h = 10
r = 3
V = 3.142 * 3^2 * 10
V = 282.78 in^3
The closest measurement is option A
Need help asap please and thank you
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Answer:
y=[tex]\frac{1}{2}[/tex]x+1
Step-by-step explanation:
y=mx+b
m is the slope of the line, which you find by counting the rise over run between two points. In this case its up one, and right two, or [tex]\frac{1}{2}[/tex].
b is the y intercept, or where the line crosses the y axis
Devon saw 19 adults wearing hats
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Answer:
please add rest of question?
Step-by-step explanation:
The con 3720bertar What can be interpreted from the youtercept of the functionRachel must pay $37 per month to use the gymRachel must pay $20 per month to use the gymRachel must pay a $37 membership fee to join the gymRachel must pay a $20 membership fee to join the samMaria wants to rent a car. She learns that the total daily costcated using the formula C = 5x + 30. hereseS driven that day. What does the constanteseer
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f(x) = 37x + 20
Answer:
Option D, $20 is the membership beause it is a fixed cost, it does not depend on the amount of months
Over a set of 5 chess games, Yolanda's rating increased 10 points, increased 4 points,
decreased 21 points, increased 23 points and decreased 8 points.
Her rating is now 1647.
What was her rating before the 5 games?
A. 1639
B. 1649
C. 1655
D. 1661
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Answer:
C. 1655
Step-by-step explanation:
+10, +4, -21, +23, -8
Adding all those terms together we get 8
1647 + 8 = 1655
Consider the line y= 3/5x-3Find the equation of the line that is parallel to this line and passes through the point (3, 4).Find the equation of the line that is perpendicular to this line and passes through the point (3, 4).
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a) y = 3/5x + 11/5
b) y = -5/3x + 9
Explanation:[tex]\begin{gathered} a)\text{ }y\text{ = }\frac{3}{5}x\text{ - 3} \\ \text{compare with equation of line:} \\ y\text{ = mx + b} \\ m\text{ =slope, b = y-intercept} \\ m\text{ =slope = 3/5} \\ b\text{ = -3} \end{gathered}[/tex]For a line to be parallel to another line. the slope of the 1st line will be equalt to the slope of the 2nd line:
slope of 1st line = 3/5
So, the slope of the 2nd line = 3/5
Given point: (3, 4) = (x, y)
To get the y-intercept of the second line, we would insert the slope and the point into the equation of line
[tex]\begin{gathered} y\text{ = mx + b} \\ 4\text{ = }\frac{3}{5}(3)\text{ + b} \\ 4\text{ = 9/5 + b} \\ 4\text{ - }\frac{\text{9}}{5}\text{ = b} \\ \frac{20-9}{5}\text{ = b} \\ b\text{ = 11/5} \end{gathered}[/tex]The equation of line parallel to y = 3/5x - 3:
[tex]\begin{gathered} y\text{ = mx + b} \\ y\text{ = }\frac{3}{5}x\text{ + }\frac{11}{5} \end{gathered}[/tex][tex]b)\text{ line perpendicular to y = 3/5x - 3}[/tex]For a line to be perpendicular to another line, the slope of one will be the negative reciprocal of the second line
Slope of the 1st line = 3/5
reciprocal of 3/5 = 5/3
negative reciprocal = -5/3
slope of the 2nd line (perpendicular) = -5/3
We need to get the y-intercept of the perpendicular line:
[tex]\begin{gathered} \text{given point: (3,4) = (x, y)} \\ y\text{ = mx + b} \\ m\text{ of the perpendicular = -5/3} \\ 4\text{ = }\frac{-5}{3}(3)\text{ + b} \\ 4\text{ = -5 + b} \\ 4\text{ + 5 = b} \\ b\text{ = 9} \end{gathered}[/tex]The equation of line perpendicular to y = 3/5x - 3:
[tex]\begin{gathered} y\text{ = mx + b} \\ y\text{ = }\frac{-5}{3}x\text{ + 9} \end{gathered}[/tex]What is a quadrilateral that has reflection symmetry, but not rotation symmetry?
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The quadrilaterals, parallelogram,square, rectangle has rotational symmetry but no reflectional symmetry
A trapezoid has neither a rotational symmetry nor a reflectional symmetry
But for an isosceles with only one pair of parallel sides has a reflectional symmetry but no rotational symmetry
Thus, the correct answer is
an isosceles with only one pair of parallel sides
complete the table of ordered pairs for the linear equation. 5x+8y=3
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Given:
5x+8y=3
The objective is to fill the table using the given values of x otr y.
Let's take that, x=0 and substitute in the given equation.
[tex]\begin{gathered} 5x+8y=3 \\ 5(0)+8y=3 \\ 0+8y=3 \\ y=\frac{3}{8} \end{gathered}[/tex]Hence, the the required solution will be (0,3/8).
Let's take that, y=0 and substitute in the given equation.
[tex]\begin{gathered} 5x+8y=3 \\ 5x+8(0)=3 \\ 5x+0=3 \\ x=3-5 \\ x=-2 \end{gathered}[/tex]Hence, the the required solution will be (-2,0).
Let's take that, y=1 and substitute in the given equation.
[tex]\begin{gathered} 5x+8y=3 \\ 5x+8(1)=3 \\ 5x+8=3 \\ 5x=3-8 \\ 5x=-5 \\ x=-\frac{5}{5} \\ x=-1 \end{gathered}[/tex]Hence, the the required solution will be (-1,1).
karen recorded her walking pace in the table below. what equation best represents this relationship
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Given Data:
The table given here shows time taken in the first column and the distance travelled in the second column.
First check the ration of distance /time to verify if the speed of the man is constant or not.
[tex]\begin{gathered} \frac{8.75\text{ m}}{2.5\text{ h}}=3.5\text{ m/h} \\ \frac{14\text{ m}}{4\text{ h}}=3.5\text{ m/h} \end{gathered}[/tex]As both ratios are same it means the speed of the man is constant and the distance travelled is directly proportional to the time taken and varies linealy with the time.
Now to determine the relationship between m and h we will use the same ratio of m and h which comes in the previous step.
[tex]\begin{gathered} \frac{m}{h}=3.5 \\ m=3.5h \end{gathered}[/tex]Thus, option (D) is the required solution.
solve the equation for x. x/16=10
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To solve further cross multiply both sides
so that
[tex]x\text{ }\times1\text{ = 16 }\times10[/tex]x = 160
Rewrite each equation in slope intercept form . Then determine whether the lines are perpendicular . Explain your answer .. y - 6 = - 5/2 (x + 4) 5y = 2x + 6
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y - 6 = - 5/2 (x + 4)
To write in slope-intercept form means to write in the form;
y= mx + b
where m is the slope and b is the intercept
y - 6 = - 5/2 (x + 4)
open the parenthesis
y - 6 = -5/2 x - 10
add 6 to both-side of the equation
y = - 5/2 x - 10 + 6
y = -5/2 x - 4
[tex]y=-\frac{5}{2}x\text{ - 4}[/tex]Next is to check whether 5y = 2x + 6 is perpendicular to the above
To do that, we have to make the equation to be in the form y=mx+ b
5y = 2x + 6
Divid through by 5
y = 2/5 x + 6/5
[tex]y\text{ = }\frac{2}{5}x\text{ + }\frac{6}{5}[/tex]The slope of perpendicular equation, when multiply gives minus one (-1)
The slope of the first equation = -5/2
The slope of the second equation is 2/5
Multiplying the two slopes;
(-5/2) (2/5) = -1
Hence the lines are perpendicular
Can someone help me with this geometry question?A.Triangular prismB.Hexagonal prismC.Triangular pyramidD.Hexagonal pyramid
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B. Hexagonal Prism
1) One prism is defined, in terms of naming it by the base.
2) Counting the edges of the base in this net surface, we can tell that this is a Hexagonal Prism for the base is a hexagon.
Given the function f(x)={4x+7 if x<0 6x+4 if x>0 _
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Given:
[tex]f(x)=\begin{cases}4x+7ifx<0{} \\ 6x+4ifx\ge0{}\end{cases}[/tex]Required:
To find the value of f(-8), f(0), f(4), and f(-100)+f(100).
Explanation:
f(-8) :
Clearly -8<0,
So
[tex]\begin{gathered} f(x)=4x+7 \\ f(-8)=4(-8)+7 \\ =-32+7 \\ =-25 \end{gathered}[/tex]f(0) :
Clearly 0=0,
[tex]\begin{gathered} f(x)=6x+4 \\ =6(0)+4 \\ =4 \end{gathered}[/tex]f(4) :
Clearly 4>0,
[tex]\begin{gathered} f(x)=6x+4 \\ f(4)=6(4)+4 \\ =24+4 \\ =28 \end{gathered}[/tex]f(-100)+f(100) :
-100<0
[tex]\begin{gathered} f(x)=4x+7 \\ f(-100)=4(-100)+7 \\ =-400+7 \\ =-393 \end{gathered}[/tex]100>0
[tex]\begin{gathered} f(x)=6x+4 \\ f(100)=6(100)+4 \\ =600+4 \\ =604 \end{gathered}[/tex][tex]\begin{gathered} f(-100)+f(100)=-393+604 \\ \\ =211 \end{gathered}[/tex]Final Answer:
[tex]\begin{gathered} f(-8)=-25 \\ \\ f(0)=4 \\ \\ f(4)=28 \\ \\ f(-100)+f(100)=211 \end{gathered}[/tex]Solve the equation 7-(5t-13)=-25-15b+21+5b=-19
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Let's solve the following expressions:
a.) 7 - (5t - 13) = -25
[tex]\text{ 7 - (5t - 13) = -25}[/tex][tex]\text{ 7 - 5t + 13 = -25}[/tex][tex]\text{ 20 - 5t = -25}[/tex][tex]\text{-5t = -25 - 20}[/tex][tex]\text{-5t = -4}5[/tex][tex]\frac{\text{-5t}}{-5}\text{ = }\frac{\text{-4}5}{-5}[/tex][tex]\text{ t = 9}[/tex]Therefore, t = 9
b.) -15b + 21 + 5b = -19
[tex]-15b+21+5b=-19[/tex][tex]-10b+21=-19[/tex][tex]-10b=-19\text{ - 21}[/tex][tex]-10b=-40[/tex][tex]\frac{-10b}{-10}=\frac{-40}{-10}[/tex][tex]\text{ b = 4}[/tex]Therefore, b = 4
What is 4x+10(2x) - 8x
Answers
4x+10(2x) - 8x
First, multiply to solve the parentheses:
4x+20x-8x
Add and subtract
16x
The endpoints are a side of a rectangle ABCD in the coordinate plane at A(3,4), B(6,1) Find the equation of the line the given segment The line segment is line Segment AB
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The endpoints are a side of a rectangle ABCD in the coordinate plane at A(3,4), B(6,1) Find the equation of the line the given segment
The line segment is line Segment AB
step 1
Find the slope of segment AB
m=(1-4)/(6-3)
m=-3/3
m=-1
step 2
Find the equation of the line in slope intercept form
y=mx+b
we have
m=-1
point (3,4)
substitute
4=(-1)*(3)+b
4=-3+b
b=4+3
b=7
therefore
the equation of segment AB is
y=-x+7Write problem as a single radical using the smallest possible root. 20
Answers
Answer::
[tex]\sqrt[30]{r^{29}}[/tex]Explanation:
Given the expression:
[tex]\sqrt[5]{r^4}\sqrt[6]{r}[/tex]First, rewrite the expression using the fractional index law:
[tex]\begin{gathered} \sqrt[n]{x}=x^{\frac{1}{n}} \\ \implies\sqrt[5]{r^4}=r^{\frac{4}{5}};\text{ and} \\ \sqrt[6]{r}=r^{\frac{1}{6}} \end{gathered}[/tex]Therefore:
[tex]\sqrt[5]{r^4}\times\sqrt[6]{r}=r^{\frac{4}{5}}\times r^{\frac{1}{6}}[/tex]Use the multiplication law of exponents:
[tex]\begin{gathered} a^x\times a^y=a^{x+y} \\ \implies r^{\frac{4}{5}}\times r^{\frac{1}{6}}=r^{\frac{4}{5}+\frac{1}{6}} \\ \frac{4}{5}+\frac{1}{6}=\frac{24+5}{30}=\frac{29}{30} \\ \operatorname{\implies}r^{\frac{4}{5}}\times r^{\frac{1}{6}}=r^{\frac{4}{5}+\frac{1}{6}}=r^{\frac{29}{30}} \end{gathered}[/tex]The resulting expression can be rewrittem further:
[tex]\begin{gathered} r^{\frac{29}{30}}=(r^{29})^{\frac{1}{30}} \\ =\sqrt[30]{r^{29}} \end{gathered}[/tex]The single radical is:
[tex]\sqrt[30]{r^{29}}[/tex]Find the missing factor. x2 - 11x + 18 = (x - 2)( .) Enter the correct answer. 000 DONE Clear all DOO
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we have the second degree polynomial
[tex]x^2-11x+18[/tex]we must find two numbers a,b such that
[tex]\begin{gathered} x^2-11x+18=(x+a)(x+b)\text{ and} \\ a+b=11 \\ ab=18 \end{gathered}[/tex]We can see that, a=-2 and b=-9 fulfill the above conditions. Therefore, we have
[tex]x^2-11x+18=(x-2)(x-9)\text{ }[/tex]write 0.751 as a percentage
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To convert decimal numbers to percentage, what we need to do is to multiply the decimal number by 100, and we will get the representation as a percentage.
In this case we have the decimal number:
[tex]0.751[/tex]We multiply that number by 100 to write is as a percentage:
[tex]0.751\times100=71.5[/tex]Answer: 75.1%
I have a question about how to solve graphing a system of inequalities and about how to do the (0,0)
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The given system of inequality is
[tex]\begin{gathered} 2x-3y>-12 \\ x+y\ge-2 \end{gathered}[/tex]At first, we must draw the lines to represent these inequalities
[tex]2x-3y=-12[/tex]Let x = 0, then find y
[tex]\begin{gathered} 2(0)-3(y)=-12 \\ 0-3y=-12 \\ -3y=-12 \\ \frac{-3y}{-3}=\frac{-12}{-3} \\ y=4 \end{gathered}[/tex]The first point is (0, 4)
Let y = 0
[tex]\begin{gathered} 2x-3(0)=-12 \\ 2x-0=-12 \\ 2x=-12 \\ \frac{2x}{2}=\frac{-12}{2} \\ x=-6 \end{gathered}[/tex]The second point is (-6, 0)
We will do the same with the second line
Let x = 0
[tex]\begin{gathered} 0+y=-2 \\ y=-2 \end{gathered}[/tex]The first point is (0, -2)
Let y = 0
[tex]\begin{gathered} x+0=-2 \\ x=-2 \end{gathered}[/tex]The second point is (-2, 0)
Since the sign of the first inequality is >, then the line will be dashed
Since the sign of the second inequality is >=, then the line will be solid
Let us substitute x, y by the origin point (0,0) in both inequalities to find the shaded part of each one
[tex]\begin{gathered} 2(0)-3(0)>-12 \\ 0-0>-12 \\ 0>-12 \end{gathered}[/tex]Since the inequality is true then the point (0, 0) lies on the shaded area
[tex]\begin{gathered} 0+0\ge-2 \\ 0\ge-2 \end{gathered}[/tex]Since the inequality is true, then point (0, 0) lies in the shaded area
Let us draw the graph
The red line represents the first inequality
The blue line represents the second inequality
The area of two colors is the area of the solution
Point (0, 0) lies in this area, then it is a solution for the given system of inequalities
The function P(x) is mapped to I(x) by a dilation in the following graph. Line p of x passes through (negative 2, 4) & (2, negative 2). Line I of X passes through (negative 4, 4) & (4, negative 2).© 2018 StrongMind. Created using GeoGebra. Which answer gives the correct transformation of P(x) to get to I(x)?
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When we're dilating a line, we can either multiply the function value by a constant
[tex]f(x)\to kf(x)[/tex]or the argument of the function
[tex]f(x)\to f(kx)[/tex]Since the y-intercept of both functions is the same, then the multiplied quantity was the argument of the function.
We want to know the constant associated to the transformation
[tex]I(x)\to I(kx)=P(x)[/tex]We have the following values for both functions
[tex]\begin{gathered} I(-4)=4,\:I(4)=-2 \\ P(-2)=4,\:P(2)=-2 \end{gathered}[/tex]For the same y-value, we have the following correlations
[tex]\begin{gathered} I(-4)=P(-2)=P(\frac{1}{2}\cdot-4) \\ I(4)=P(2)=P(\frac{1}{2}\cdot4) \\ \implies I(x)=P(\frac{1}{2}x) \end{gathered}[/tex]and this is our answer.
[tex]I(x)=P(\frac{1}{2}x)[/tex]A training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 96 m long and 64 m wide. Find the area of the training field. Use the value 3.14 for n, and do not round your answer. Be sure to include the correct unit in your answer.
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To find:
The area of the training field.
Solution:
The training field is made of two semicircles and a rectangle.
The length and width of the rectangle is 96 m and 64 m. So, the area of the rectangle is:
[tex]\begin{gathered} A=l\times w \\ =96\times64 \\ =6144\text{ m}^2 \end{gathered}[/tex]The diameter of the semicircle is 64 m. SO, the radius of the semicircle is 32 m.
The area of two semicircles is:
[tex]\begin{gathered} A=2\times\frac{1}{2}\pi r^2 \\ =3.14\times(32)^2 \\ =3.14\times1024 \\ =3215.36 \end{gathered}[/tex]So, the area of the training field is:
[tex]\begin{gathered} A=6144+3215.36 \\ =9359.36 \end{gathered}[/tex]Thus, the area of the training field is 9359.36 m^2.
I need help figuring out how to write out this problem correctly.
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Answer
[tex]\frac{\sqrt{10}}{11}[/tex]Step-by-step explanation
Given the expression:
[tex]\sqrt{\frac{10}{121}}[/tex]Distributing the square root over the division and evaluating the square root at the denominator:
[tex]\begin{gathered} \frac{\sqrt{10}}{\sqrt{121}} \\ \frac{\sqrt{10}}{11} \end{gathered}[/tex]Find the x-intercepts and y-intercept of the following
function.
f(x) = (x+7) (x + 1)(x − 2)
Write your answer in coordinate pairs of the form (x, y).
Provide your answer below:
x-intercept: ().(
]).() and
y-intercept:
Answers
Step-by-step explanation:
there are 3 x-intercepts (the x- values when y = 0). because y is only 0, when one of the 3 factors is 0.
and that is the case for
x = -7
x = -1
x = 2
so, formally, the x- intercepts are
(-7, 0), (-1, 0), (2, 0)
the y intercept is the y- value when x = 0.
(0 + 7)(0 + 1)(0 - 2) = 7×1×-2 = -14
the y-intercept is formally
(0, -14)
Conner plans to plow a field in one day. Before lunch he plows 15 acres, which is 30% of the field. Howmany acres will he have to plow after lunch in order to finish the field?
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Solution:
Let the total field to plow be 100 %
According to the question,
The field plowed before lunch is shown below:
15 acres field = 30%
30 % field = 15 acres
1 % field = 15/30 acres
The field plow after lunch is 70%.
[tex]\begin{gathered} 70\text{ \% of field = }\frac{15}{30}\times70 \\ =35\text{ acres} \end{gathered}[/tex]Final Answer:
Therefore, the field to plow after lunch in order to finish the field is 35 acres.
1 4 a) Is the above sequence arithmetic? Justify your answer. b) Write the explicit formula for the above sequence. c) Find the 18th term.
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THe length of each cube is inreasing by 1. This means that the increment is in arithmetic sequence.
The formula for determining the nth term of an aritmetic sequence is expressed as
Tn = a + (n - 1)d
Where
a represents the first tem of the sequence
Common difference, d represents the difference between consecutive terms
Given that a = 3, d = 1
Tn = 2 + (n - 1)1
Tn = 2 + n - 1
Tn = 1 + n
For the 18th term, n = 18
T18 = 1 + 18 = 19
The number of boxes in the square for the 19th term is
19 * 19 = 361
You are scuba diving at 120 feet below sea level. You begin to ascend at a rate of 4 feet per second.a. Where will you be 10 seconds after you begin your ascension? b. How long will it take to reach the surface?
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The ascension can be modeled using the function:
[tex]d(t)=d_0-r\cdot t[/tex]Where d is the number of feet below the sea level at time t (in seconds), d₀ is the initial "depth", and r is the ascension rate.
From the problem, we identify:
[tex]\begin{gathered} r=4\text{ feet per second} \\ d_0=120\text{ feet} \end{gathered}[/tex]Then:
[tex]d(t)=120-4t[/tex]a)
After 10 seconds, we have t = 10:
[tex]\begin{gathered} d(10)=120-4\cdot10=120-40 \\ \\ \Rightarrow d(10)=80\text{ feet} \end{gathered}[/tex]After 10 seconds, we will be 80 feet below sea level.
b)
To find how long will it take to reach the surface, we need to solve the equation d(t) = 0.
[tex]\begin{gathered} d(t)=0 \\ 120-4t=0 \\ 4t=120 \\ \\ \therefore t=30\text{ seconds} \end{gathered}[/tex]We will reach the surface after 30 seconds.
A local band was interested in the average song time for rock bands in the 1990s. They sampled eight different rock bands and found that the average time was 3.19 minutes with a standard deviation of 0.77 minutes.
Calculate the 95% confidence interval (in minutes) for the population mean.
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The 95% confidence interval (in minutes) for the population mean is of:
(2.55, 3.83).
What is a t-distribution confidence interval?The bounds of the confidence interval are given according to the following rule:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
In which the parameters are described as follows:
[tex]\overline{x}[/tex] is the sample mean.t is the critical value.n is the sample size.s is the standard deviation for the sample.The distribution is used when the standard deviation of the population is not known, only for the sample.
In the context of this problem, the values of the parameters are given as follows:
[tex]\overline{x} = 3.19, s = 0.77, n = 8[/tex]
The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 8 - 1 = 63 df, is t = 2.3646.
Then the lower bound of the confidence interval is calculated as follows:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 3.19 - 2.3646\frac{0.77}{\sqrt{8}} = 2.55[/tex]
The upper bound is calculated as follows:
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 3.19 + 2.3646\frac{0.77}{\sqrt{8}} = 3.83[/tex]
More can be learned about the t-distribution at https://brainly.com/question/16162795
#SPJ1
State whether the given set of lines are parallel, perpendicular or neither.3x-2y=56y-9x=6The lines are Answer
Answers
Two lines are parallel if:
[tex]m1=m2[/tex]Two lines are perpendicular if:
[tex]m1\cdot m2=-1[/tex]---------------------
Let's rewrite the given equations in the slope-intercept form:
[tex]\begin{gathered} 3x-2y=5 \\ y=\frac{3}{2}x-\frac{5}{2} \\ -------- \\ 6y-9x=6 \\ y=\frac{3}{2}x+1 \end{gathered}[/tex]Since:
[tex]\begin{gathered} m1=m2 \\ \frac{3}{2}=\frac{3}{2} \\ \end{gathered}[/tex]We can conclude that the lines are parallel.