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3 years ago

6

Please help find the anwser to question 1 and 2

1 answer:

grin007 [14]3 years ago

4 0

16ft and 25ft sorry if i’m wrong

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1) 2(a + 3) = -12

serg [7]

1)
2(a + 3) = -12
Divide by 2
a + 3 = -6
Subtract 3
a = -9

2)
3(p + 2) = 18
Divide by 3
p + 2 = 6
Subtract 2
p = 4

3)
4(2r + 8) = 88
Dive by 4
2r + 8 = 22
Subtract 8
2r = 14
Divide by 2
r = 7

4)
2(3a + 2) = -8
Divide by 2
3a + 2 = -4
Subtract 2
3a = -6
Divide by 3
a = -2

5)
4(k + 3) = 4
Divide by 4
k + 3 = 1
Subtract 3
k = -2

3 0

3 years ago

Read 2 more answers

If n is the least of two consecutive odd integers, which of the following represents the sum of the two integers?

Airida [17]

Let the first odd integer = n
∴ The second <span>consecutive odd integer = n+2

∴ </span><span>The sum of the two integers = (n) + (n+2)
= 2n + 2

</span> The correct choice is option (D)
<span> D) 2n + 2 </span>

8 0

3 years ago

raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the ci

andreyandreev [35.5K]

Answer:

Equation:

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.

Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

The center of this circle is the midpoint of (-3, 4) and (5, -2).

We use the midpoint formula:

$( \frac{x_1+x_2}{2}, \frac{y_1+y_2,}{2} )$

Plug in the points to get:

$( \frac{ - 3+5}{2}, \frac{ - 2+4}{2} )$

$( \frac{ -2}{2}, \frac{ 2}{2} )$

$( - 1, 1)$

We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:

$r = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$r = \sqrt{ {(5 - - 1)}^{2} + {( - 2- - 1)}^{2} }$

$r = \sqrt{ {(6)}^{2} + {( - 1)}^{2} }$

$r = \sqrt{ 36+ 1 } = \sqrt{37}$

The equation of the circle is given by:

$(x-h)^2 + (y-k)^2 = {r}^{2}$

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

$(x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2}$

$(x + 1)^2 + (y - 1)^2 = 37$

We expand to get:

${x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37$

${x}^{2} + {y}^{2} + 2x - 2y +2 - 37= 0$

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

${x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0$

${x}^{2} +2x - 35= 0$

$(x - 5)(x + 7) = 0$

$x = 5 \: or \: x = - 7$

The point (5,0) and (-7,0) lies on the circle.

When x=0

${0}^{2} + {y}^{2} + 2(0) - 2y - 35= 0$

${y}^{2} - 2y - 35= 0$

$(y - 7)(y + 5) = 0$

$y = 7 \: or \: y = - 5$

The point (0,-5) and (0,7) lie on this circle.

3 0

3 years ago

List the interior angles shown in the figure

Oksi-84 [34.3K]

Answer:

The angles are 5,2, and 3.

Step-by-step explanation:

Well, they're inside the triangle.

3 0

3 years ago

Read 2 more answers

HELP ME 35 1/4 + 12 7/12

ZanzabumX [31]

The answer to this question is 48.

7 0

4 years ago