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3 years ago

identifiy an equation in point slope form for the line perpendicular to y=3x+5 that passes through (4,-1)

Mathematics

1 answer:

dedylja [7]3 years ago

6 0

Answer:

Hey there!

A line perpendicular to y=3x+5 would have a slope of -1/3.

Thus, the problem now becomes to write a line in point slope form with a slope of -1/3, and that passes through (4, -1)

y+1=-1/3(x-4), would be your answer.

Hope this helps :)

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The function h(t) = −5t2 + 20t shown in the graph models the curvature of a satellite dish: graph of parabola starting at zero c

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Answer:

Part 1) Option B: 0 ≤ x ≤ 4

Part 2) Option C: (5x − 1)(2x − 3)

Part 3) Option C: The crop yield decreased by 10 pounds per acre from year 10 to year 20.

Part 4) Option B: f(x) = (2x + 1)(2x − 1)

Part 5) Option A: The company earned 85 billion dollars in 2008

Part 6) Option C: (3, 0) and (5, 0)

Step-by-step explanation:

Part 1) we have

$h(t)=-5t^{2}+20t$

This is a vertical parabola open downward

The vertex is a maximum

The x-intercepts are the points (0,0) and (4,0)

The vertex is the point (2,20) ---> The x-coordinate of the vertex is the midpoint of the x-intercepts

The domain is the interval ----> [0,4]

$0\leq t\leq 4$

The range is the interval ----> [0,20]

$0\leq h(t)\leq 20$

Part 2) we have

$V=10x^{2} -17x+13$

where

x is the time in minutes

V is the volume in gallons

we know that

When the trough is empty, the volume is equal to zero

For V=0

$10x^{2} -17x+3=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$10x^{2} -17x+3=0$

$a=10\\b=-17\\c=3$

substitute in the formula

$x=\frac{-(-17)(+/-)\sqrt{-17^{2}-4(10)(3)}} {2(10)}$

$x=\frac{17(+/-)\sqrt{169}} {20}$

$x=\frac{17(+/-)13} {20}$

$x=\frac{17(+)13} {20}=\frac{3}{2}$

$x=\frac{17(-)13} {20}=\frac{1}{5}$

$10x^{2} -17x+3=10(x-\frac{3}{2})(x-\frac{1}{5})$

simplify

$10(x-\frac{3}{2})(x-\frac{1}{5})=(2x-3)(5x-1)$

Part 3) we have

$f(x)=-x^{2}+20x+100$

we know that

To find the average rate of change, we divide the change in the output value by the change in the input value

the average rate of change is equal to

$\frac{f(b)-f(a)}{b-a}$

In this problem we have

$f(a)=f(10)=-(10)^{2}+20(10)+100=200$

$f(b)=f(20)=-(20)^{2}+20(20)+100=100$

$a=10$

$b=20$

Substitute

$\frac{100-200}{20-10}$

$-\frac{100}{10}=-10$ ----> is a decreasing function

therefore

The crop yield decreased by 10 pounds per acre from year 10 to year 20.

Part 4) we have

$f(x)=4x^{2} -1$

<em>Find out the x-intercepts</em>

Remember that the x-intercepts are the values of x when the value of the function is equal to zero

For f(x)=0

$4x^{2} -1=0$

Solve for x

$4x^{2}=1$

$x^{2}=\frac{1}{4}$

take square root both sides

$x=(+/-)\frac{1}{2}$

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The expression represent a company's net sales, in billions, from 2008 to 2018

t is the number of years since the end of 2008

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$(0)^{2}+14(0)+85=85\ billion\ dollars$

therefore

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Part 6) What are the x-intercepts of the parabola?

we have the points (2,3),(4,-1),(6,3)

This is a vertical parabola open upward

The vertex represent a minimum

In this problem the vertex is the point (4,-1)

The equation of a vertical parabola in vertex form is equal to

$y=a(x-h)^2+k$

where

a is a coefficient

(h,k) is the vertex

we have

(h,k)=(4,-1)

substitute

$y=a(x-4)^2-1$

take the point (2,3) and substitute in the quadratic equation to solve for a

For x=2, y=3

$3=a(2-4)^2-1$

$3=4a-1$

$4a=4$

$a=1$

The quadratic equation is

$y=(x-4)^2-1$

Find out the x-intercepts

Remember that the x-intercepts are the values of x when the value of the function is equal to zero

For y=0

$(x-4)^2-1=0$

Solve for x

$(x-4)^2=1$

take square root both sides

$(x-4)=(+/-)1$

$x=4(+/-)1$

$x=4+1=5\\x=4-1=3$

therefore

The x-intercepts are the points (3,0) and (5,0)

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