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stira [4]
2 years ago
5

I need help anyone! Fill in the blank: what reason belongs in step 10?

Mathematics
1 answer:
emmasim [6.3K]2 years ago
4 0

I think 2 is the isosceles triangle theorem.

I think 10 is SAS

11 is CPCTC (My favorite thing to say)

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Three consecutive even integers whose sum is -48​
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Three consecutive even integers whose sum is -48
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Cheyenne drove 3000 miles in 75 hours. At this rate, how long will it take her to drive 4000 miles?
lyudmila [28]
Set up a proportion:
\frac{3000}{75}=\frac{4000}{x}
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Can someone please help me with this
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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
3 years ago
Amy, Bess, Cassie, Diana, and Ellen are a basketball team. Two of them will be guards. Assuming they all have an equal chance of
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