HELPPP MEEE PLEASEEEEE

Evaluate the following equation for x. 2/5x = −20

kenny6666 [7]

2/5x=-20
x=-50
Explanation:
Whenever you take the 2/5 and divide it from -20 you are left with x=-50. Therefore, the value for x is -50.

6 0

3 years ago

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

Lucy buys baseball cards in packages of 10. List the number

Vlada [557]

Answer:

Write the following numbers in order from smallest value to largest value: 1.75, 1. Justify your answer. 4. If 2x2 – x + 6 is subtracted from x2 + 3x – 2, ...

Step-by-step explanation:

1. Which number is irrational?

(A) (B) (C) 0.3333 (D)

2. In which group are the numbers arranged in order from smallest value to largest value?

(A) π, 3.14, (B) 3.14, π

(C) , 3.14, π, (D) 3.14, , π,

3. Write the following numbers in order from smallest value to largest value:

1.75, 1

Justify your answer.

4. If 2x2 – x + 6 is subtracted from x2 + 3x – 2, the result is

(A) x2 + 2x – 8 (B) x2 – 4x + 8 (C) –x

2 + 2x – 8 (D) –x

2 + 4x – 8

7 0

2 years ago

Which steps can be used to solve for the value of 'y'? 2/3(y+57)=178

bagirrra123 [75]

2 / 3 ( y + 57 ) = 178

2( y + 57 ) / 3 = 178

2*y + 2 * 57 / 3 = 178

2y + 114 / 3 = 178

3*(2 y + 114 ) / 3 * ( 178)

2 y + 114 = 534

2y = 534 - 114

2 y = 420

y = 420 / 2

y = 210

hope this helps!

3 0

3 years ago

In the year 2009, a company made $8.4 million in profit. For each consecutive year

Elenna [48]

Answer:

$10.8 million or 10,752,000

Step-by-step explanation:

14% of 8,400,000 is 1,176,000

t = time in years

In one year, the company would make 8,400,000+(t*1,176,000)

In 2 years they make 8,400,000+(2*1,176,000) = 10,752,000

now round 10,752,000 to the nearest hundred thousand, which is 1/10 of 1,000,000.

3 0

3 years ago