When using estimates to make decisions,

Write the standard form of the line that passes through the given points. Include your work in your final answer. Type

bekas [8.4K]

9514 1404 393

Answer:

5x -3y = 44

Step-by-step explanation:

We choose to start with the form ...

(Δy)x -(Δx)y = (Δy)(x1) -(Δx)(y1) . . . . . where Δy = y2 -y1 and Δx = x2 -x1

For this purpose, it is convenient to choose y2 > y1, so we have ...

Δy = -3 -(-8) = 5

Δx = 7 -4 = 3

5x -3y = 5(4) -3(-8)

5x -3y = 44

7 0

3 years ago

Find the average squared distance between the points of r{(x,y): 0x, 0y} and the point (,)

ivolga24 [154]

The average squared distance between the points is their variance

The average squared distance is 8/3

<h3>How to determine the average squared distance?</h3>

The given parameters are:

R={(x,y): 0<=x<=2, 0<=y<=2}

The point = (2,2).

f(x,y) = (x-2)² + (y-2)²

The squared distance is calculated as:

D² = f(x,y) = (x-2)² + (y-2)²

Where the area (A) is:

A = xy

Substitute the maximum values of x and y in A = xy

A = 2 * 2

A = 4

The minimum values of x and y are 0.

So, the limits for the integrals are 0 to 2

The integral becomes

$D\² = \int \int (x-2)\² + (y-2)\² dx dy$

Expand

$D\² = \int \int x\² -4x + 4 + (y-2)\² dx dy$

Evaluate the inner integral with respect to x from 0 to 2

$D\² = \int \frac 13x^3 -2x^2 + 4x + x(y-2)\² |\limits^2_0 dy$

Expand

$D\² = \int [\frac 13(2)^3 -2(2)^2 + 4(2) + (2)(y-2)\²] dy$

Simplify

$D\² = \int \frac 83 + (2)(y-2)\² dy$

Expand

$D\² = \int \frac 83 + 2y^2-8y + 8 \ dy$

Evaluate the integral with respect to y from 0 to 2

$D\² = [\frac 83y + \frac 23y^3- 4y^2 + 8y ]|\limits^2_0$

Expand

$D\² = \frac 83*2 + \frac 23*2^3- 4*2^2 + 8*2$

D² = 32/3

The average squared distance (AD²) is calculated as:

AD² = D²/A

So, we have:

AD² = 32/3 $\div$ 4

Evaluate the quotient

AD² = 32/12

Simplify

AD² = 8/3

Hence, the average squared distance is 8/3

Read more about average distance or variance at:

brainly.com/question/15858152

8 0

3 years ago

Jeb is the local tennis pro at the country club, and he needs to know at what height the tennis ball needs to be when he hits it

suter [353]

Answer:

The height from ground, at which Jeb needs to hit the ball is 7.8 meters

Step-by-step explanation:

The given parameters are;

The distance away the tennis ball is hit = 20 meters;

The height of the tennis net = 0.3 m

The distance from the net the ball lands = 4 meters

The time the ball takes to land from the 0.3 m net height is given as follows;

t = √(h/(1/2×g)) = √(0.3/(1/2 × 9.8)) ≈ 0.2474

t ≈ 0.2474 seconds

The speed of the ball = Distance/Time = 4/0.2474 ≈ 16.166

The speed of the ball ≈ 16.166 m/s

The time it takes the ball to reach the net = 20/16.166 ≈ 1.24

The time it takes the ball to reach the net, tₓ ≈ 1.24 seconds

The height, hₓ the ball falls in 1.24 seconds is given as follows;

hₓ = 1/2·g·tₓ² = 1/2 × 9.8 × 1.24² = 7.5

The height, hₓ the ball falls in 1.24 seconds = 7.5 m

The height the ball falls in the time taken to reach the net from 20 meters = The height, hₓ the ball falls in 1.24 seconds

Therefore, the height at which Jeb needs to hit the ball = The height the ball falls in the time taken to reach the net from 20 meters + The height of the net = 7.5 m + 0.3 m = 7.8 m

The height at which Jeb needs to hit the ball = 7.8 m

4 0

3 years ago

My square orchid garden abuts my house so that the house itself forms the northern boundary. The fencing for the southern bounda

lilavasa [31]

Answer:

The total cost of the fencing as a function of side is $C = 8\cdot x$ .

Step-by-step explanation:

Dimensionally speaking, the total cost is equal to the product of length and cost per length. Given that orchid garden has a square form, there is only an input: Side ( $x$ ), measured in feet. Therefore, the total cost of fencing is represented by the following formula: