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3 years ago

5

81÷3 using distributive property

2 answers:

shepuryov [24]3 years ago

7 0

81 divided by 3 equals 27

Brums [2.3K]3 years ago

6 0

I hope this helps you

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Find the equation of the line. Use exact numbers <br> y= ?x+ ?

lina2011 [118]

Answer:

y=3x+3

Step-by-step explanation:

slope is 3 because rise over run. Rise is 3 and run is 1. 3/1 is 3. and b= y-intercept and the poiunt wehre the line intercepts the y axis is 3. so your answer is y=3x+3

8 0

3 years ago

Read 2 more answers

Two lines are perpendicular, and neither line is horizontal. Which of the following is not necessarily true of the lines?

musickatia [10]

I think the correct answer from the choices listed above is option C. <span>Two lines are perpendicular, and neither line is horizontal so it must be that the slopes of the lines are opposite reciprocals, one slope is positive and the other has a negative slope. Having different slopes so they should intersect the y-axis at different values.</span>

7 0

3 years ago

Write the standard form of the equation of the circle with center (−7,1) that passes through the point (−7,9).

kiruha [24]

Answer:

(x + 7)² + (y - 1)² = 64

Step-by-step explanation:

Use the equation of a circle, (x - h)² + (y - k)² = r²

Plug in the point and center, then solve for r:

(x - h)² + (y - k)² = r²

(-7 + 7)² + (9 - 1)² = r²

0 + 64 = r²

64 = r²

8 = r

Then, plug in the center and r² into the equation:

(x + 7)² + (y - 1)² = 64

So, (x + 7)² + (y - 1)² = 64 is the standard form of the circle's equation

4 0

3 years ago

Find the distance between -3 and 4.<br><br> A) -12 <br> B) -7 <br> C) -1 <br> D) 7

bonufazy [111]

The answer is 7 because add them to see the distance since it’s both positive if you count the distance it is away from zero so 3+4=7

7 0

3 years ago

Evaluate the double integral. 2y2 dA, D is the triangular region with vertices (0, 1), (1, 2), (4, 1) D

zubka84 [21]

Answer:

$\mathbf{\iint _D y^2 dA= \dfrac{22}{3}}$

Step-by-step explanation:

From the image attached below;

We need to calculate the limits of x and y to find the double integral

We will notice that y varies from 1 to 2

The line equation for (0,1),(1,2) is:

$y-1 = \dfrac{2-1}{1-0}(x-0)$

y - 1 = x

The line equtaion for (1,2),(4,1) is:

$y-2 = \dfrac{1-2}{4-1}(x-1) \\ \\ y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = (x -1)

-3y + 6 = x - 1

-x = 3y - 6 - 1

-x = 3y - 7

x = -3y + 7

This implies that x varies from y - 1 to -3y + 7

Now, the region D = {(x,y) | 1 ≤ y ≤ 2, y - 1 ≤ x ≤ -3y + 7}

The double integral can now be calculated as:

$\iint _D y^2 dA= \int ^2_1 \int ^{-3y +7}_{y-1} \ 2y ^2 \ dx \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[ 2xy ^2 \bigg]^{-3y+7}_{y-1} \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[2(-3y+7)y^2-2(y-1)y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[-6y^3 +14y^2 -2y^3 +2y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[-8y^3 +16y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \bigg[-8(\dfrac{y^4}{4}) +16(\dfrac{y^3}{3})\bigg ] ^2_1$

$\iint _D y^2 dA= \bigg[-8(\dfrac{16}{4}-\dfrac{1}{4}) +16(\dfrac{8}{3}-\dfrac{1}{3})\bigg ]$

$\iint _D y^2 dA= \bigg[-8(\dfrac{15}{4}) +16(\dfrac{7}{3})\bigg ]$

$\iint _D y^2 dA= -30 + \dfrac{112}{3}$

$\iint _D y^2 dA= \dfrac{-90+112}{3}$

$\mathbf{\iint _D y^2 dA= \dfrac{22}{3}}$

4 0

3 years ago