Ask question

Ask question

All categories

2 years ago

6

Divide the following polynomial, then place the answer in the proper location on the grid. (9x 4 + 3x 3y - 5x 2y 2 + xy 3) / (3x

3 + 2x 2y - xy 2)

1 answer:

olga55 [171]2 years ago

3 0

The answer is -4+72/2y+9 when you solve using polynomial division :)

You might be interested in

Find the value for β that makes cosβ = sin15 true.

Gemiola [76]

Answer:

Step-by-step explanation:

This is an equation for which you need to solve for the value of the missing angle beta.

$cos\beta =sin(15)$

sin(15) has a decimal answer that you will find on your calculator. Finding that value makes this equation then

cosβ = .2588190451

Now use the 2nd button and then the cos button to find the angle that has a cos of that decimal. When you hit 2nd then cos, you should see this on your screen:

cos⁻¹ (

After that open parenthesis, enter you decimal then hit the enter button to find the angle. You should get 75, which is Choice B.

8 0

3 years ago

Bla nnn,,,,,,bs jfbjnf msnjkmfe sjnmf cmmnefnmv fcve

EleoNora [17]

Answer:

Ummmmm ehhhhhh

8 0

2 years ago

Read 2 more answers

Pentagon ABCDE was enlarged: a What is the ratio of enlargement to the original? Convert the ratio to a percent.

Step2247 [10]

Answer:

hello, yes or nou sorry jajaj

6 0

2 years ago

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

Triangles practice, need help.

Sliva [168]

Y is 15 and x is 14 mark me as brainless

8 0

2 years ago

Read 2 more answers