Finish the pattern: 3,6,12,__,_

Can someone please help me with this

Mama L [17]

Answer:

27.

Equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1: 400 feet

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

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

Part 4: t = 10 seconds

29.

Part 1: Average Rate of Change = -15

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

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

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

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

y = -7x + 1

y = x + 1

28.

Part 1:

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

Part 2:

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

Part 3:

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

Part 4:

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.

Part 1:

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 = $\frac{-13-2}{3-2}=-15$

Part 2:

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

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B25-10%5Csqrt%7B3%7D%2B3%20%7D" id="TexFormula1" title="\sqrt{25-10\sqrt{3}+3 }" alt=

Varvara68 [4.7K]

Here is your answer: )

6 0

3 years ago

What is the slope of the line on the graph? Enter your answer in the box

miskamm [114]

Answer:

The slope is 6

Step-by-step explanation:

Rise/run

The rise is 6 and the run 1

Then simplfy 6/1 to get 6

5 0

3 years ago

Please help explain the steps

marusya05 [52]

Answer:

the factored equation is ((x^3) + 3x^2) + 2x = 0

Step-by-step explanation:

*The ^ means to the power of.

so you should start off by splitting it into ways, like isolate 2x and start with x^3 and 3x^2, if thats the wrong answer u are not looking for then the other formula is x^3 + 3x^2 + 2x = x • (x^2 + 3x + 2), again just splitting it into ways. since we know the equation is = to 0, we put 2 of them together, and since you can divide everything by x and just put x outside to keep it the same, it would become x^3 + 3x^2 + 2x = x • (x^2 + 3x + 2).

Your Welcome

6 0

4 years ago

Emily and Molly are selling pies for a school fundraiser. Customers can buy cherry pies and lemon pies. Emily sold 4 cherry pies

Natalka [10]

Answer:

<h2>cherry pie = 5</h2><h2>lemon pie =10</h2>

Step-by-step explanation:

Step one:

let cherry be x

and lemon be y

the systems of the equation for the solution is given as

4x+y=30-----------1

8x+6y=100------2

Step two:

solve the equation simultaneously by substitution we have

4x+y=30

y=30-4x

put y=30-4x in equation 2

8x+6(30-4x)=100

8x+180-24x=100

8x-24x=100-180

-16x=-80

x=80/16

x=5

put x=4 in y=30-4x

y=30-4(5).

y=30-20

y=10

5 0

3 years ago

Finish the pattern: 3,6,12,__,__

Finish the pattern: 3,6,12,,