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Paladinen [302]

2 years ago

13

Select the correct answer. If the graph of function g is 6 units below the graph of function f, which could be function g? f(x)

= -2x + 7
A. g(x) = -2x − 6
B. g(x) = -2x + 20
C. g(x) = -6x + 7
D. g(x) = -2x + 1

1 answer:

Vitek1552 [10]2 years ago

5 0

D. g(x) = -2x + 1
this is because if you wanted the function to be six below, the slope would not change but the y-intercept would go down 6

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A grocery store sells a bag of 8 oranges for $4.24. What is the unit cost?

NNADVOKAT [17]

Answer:

0.53

Step-by-step explanation:

8/8 = 1

4.24/8 = 0.53

1/0.53 = 0.53

Answer: 0.53 :D

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2 years ago

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Can someone please help me with this

Mama L [17]

Answer:

27.

<em>Equation of 2 tangent lines at the given curve going through point P is:</em>

<em>y = -7x + 1</em>

<em>y = x + 1</em>

28.

<u>Part 1:</u> 400 feet

<u>Part 2:</u> Velocity is +96 feet/second (or 96 feet/second UPWARD) & Speed is 96 feet per second

<u>Part 3:</u> acceleration at any time t is -32 feet/second squared

<u>Part 4:</u> t = 10 seconds

29.

<u>Part 1:</u> Average Rate of Change = -15

<u>Part 2:</u> The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

<u />

Step-by-step explanation:

27.

First of all, the equation of tangent line is given by:

$y-y_1=m(x-x_1)$

Where m is the slope, or the derivative of the function

Now,

If we take a point x, the corresponding y point would be $x^2-3x+5$ , so the point would be $(x,x^2-3x+5)$

Also, the derivative is:

$f(x)=x^2-3x+5\\f'(x)=2x-3$

Hence, we can equate the DERIVATIVE (slope) and the slope expression through the point given (0,1) and the point we found $(x,x^2-3x+5)$

The slope is $\frac{y_2-y_1}{x_2-x_1}$

So we have:

$\frac{x^2-3x+5-1}{x-0}\\=\frac{x^2-3x+4}{x}$

Now, we equate:

$2x-3=\frac{x^2-3x+4}{x}$

We need to solve this for x. Shown below:

$2x-3=\frac{x^2-3x+4}{x}\\x(2x-3)=x^2-3x+4\\2x^2-3x=x^2-3x+4\\x^2=4\\x=-2,2$

So, this is the x values of the point of tangency. We evaluate the derivative at these 2 points, respectively.

$f'(x)=2x-3\\f'(-2)=2(-2)-3=-7\\f'(2)=2(2)-3=1$

Now, we find 2 equations of tangent lines through the point (0,1) and with respective slopes of -7 and 1. Shown below:

$y-y_1=m(x-x_1)\\y-1=-7(x-0)\\y-1=-7x\\y=-7x+1$

and

$y-y_1=m(x-x_1)\\y-1=1(x-0)\\y-1=x\\y=x+1$

<em>So equation of 2 tangent lines at the given curve going through point P is:</em>

<em>y = -7x + 1</em>

<em>y = x + 1</em>

<em></em>

28.

<u>Part 1:</u>

The highest point is basically the maximum value of the position function. To get maximum, we differentiate and set it equal to 0. Let's do this:

$s(t)=160t-16t^2\\s'(t)=160-32t\\s'(t)=0\\160-32t=0\\32t=160\\t=\frac{160}{32}\\t=5$

So, at t = 5, it reaches max height. We plug in t = 5 into position equation to find max height:

$s(t)=160t-16t^2\\s(5)=160(5)-16(5^2)\\=400$

max height = 400 feet

<u>Part 2:</u>

Velocity is speed, but with direction.

We also know the position function differentiated, is the velocity function.

Let's first find time(s) when position is at 256 feet. So we set position function to 256 and find t:

$s(t)=160t-16t^2\\256=160t-16t^2\\16t^2-160t+256=0\\t^2-10t+16=0\\(t-2)(t-8)=0\\t=2,8$

At t = 2, the velocity is:

$s'(t)=v(t)=160-32t\\v(2)=160-32(2)\\v(2)=96$

It is going UPWARD at this point, so the velocity is +96 feet/second or 96 feet/second going UPWARD

The corresponding speed (without +, -, direction) is simply 96 feet/second

<u>Part 3:</u>

We know the acceleration is the differentiation of the velocity function. let's find it:

$v(t)=160-32t\\v'(t)=a(t)=-32$

hence, the acceleration at any time t is -32 feet/second squared

<u>Part 4:</u>

The rock hits the ground when the position is 0 (at ground). So we equate the position function, s(t), to 0 and find time when it hits the ground. Shown below:

$s(t)=160t-16t^2\\0=160t-16t^2\\16t^2-160t=0\\16t(t-10)=0\\t=0,10$

We disregard t = 0 because that's basically starting. So we take t = 10 seconds as our answer and we know rock hits the ground at t = 10 seconds.

29.

<u>Part 1:</u>

The average rate of change is basically the slope, which is

Slope = Change in y/ Change in x

The x values are given, from 2 to 3, and we need to find corresponding y values by plugging in the x values in the function. So,

When x = 2, $y=f(2)=-(2)^3 + 4(2) + 2=2$

When x = 3, $y=f(3)=-(3)^3 + 4(3) + 2=-13$

Hence,

Average Rate of Change = $\frac{-13-2}{3-2}=-15$

<u>Part 2:</u>

The instantaneous rate of change is got by differentiating the function and plugging the 2 points and finding the difference.

First, let's differentiate:

$f(x)=-x^3+4x+2\\f'(x)=-3x^2+4$

Now, find the derivative at 3,

$f'(x)=-3x^2+4\\f'(3)=-3(3)^2+4=-23$

finding derivative at 2,

$f'(x)=-3x^2+4\\f'(2)=-3(2)^2+4=-8$

The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

6 0

2 years ago

10 points!! Will mark brainliest!!! Please help me I don’t understand this very well

Katen [24]

Answer:

9?

Step-by-step explanation:

4 0

3 years ago

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Janis uses the equation C = 4h + 8 to find the total cost, C, in dollars, of renting a bike for h hours. If Janis does not spend

Digiron [165]

Answer: she can rent the bike for a maximum of 13 hours.

Step-by-step explanation:

Let h represent the number of hours for which she rents the bike.

The equation used by Janis to find the total cost, C, in dollars, of renting a bike for h hours is expressed as

C = 4h + 8

If Janis does not spend more than $60, then it means that

4h + 8 ≤ 60

4h ≤ 60 - 8

4h ≤ 52

h ≤ 52/4

h ≤ 13

Therefore, the maximum number of hours for which she can rent the bike is 13.

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3 years ago

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bekas [8.4K]

Point form: (2,3) and x=2 and y=3

4 0

2 years ago