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Alina [70]
3 years ago
6

Select all that apply

Mathematics
1 answer:
notsponge [240]3 years ago
6 0

Answer:D

Step-by-step explanation:

2x + 3y = 12

-2x -2x

3y = -2x + 12

/3 /3 /3

Y = -2/3x + 4

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What is the slope of the line that passes through the points (7.-12) and (-9,36)?​
REY [17]

Answer:

Step-by-step explanation:

m = slope where

m = rise / run

rise = y2 - y1  

run = x2 - x1

where  

the given point P1 = (7, -12)

and is in the form of (x1,y1)  

and

the given point P2 = (-9,36)  

is  in   the   form  of   (x2,y2)

then

m =  ( y2 - y1 ) / ( x2 - x1 )

m = ( 36 - (-12) ) / ( -9 -7 )

m = ( 36 + 12 ) /  ( - 16 )

m =  48 /  - 16

m = - 3

the slope is negative 3

slope = - 3

4 0
2 years ago
What is the volume of a right circular cylinder with a base diameter of 6 m and a height of 5 m?
ikadub [295]
Radius = 3 m
Cylinder Volume =<span>  </span><span>π <span>• r² • height
</span></span>
Cylinder Volume =<span>  3.14159 * 3^2 * 5

</span><span><span><span>Cylinder Volume =<span>  </span>141.372

</span></span></span>OR  PI * 45


6 0
3 years ago
Two angles are supplementary if the sum of their measure is 180 degrees. If one of the two supplementary angles measures 85 degr
ale4655 [162]

Answer:

95 degrees.

Step-by-step explanation:

Their sum is 180. You will subtract 85 to get the other half of the whole.

180 - 85 = 95

The other angle is 95 degrees.

4 0
3 years ago
The slope of a line is . What is the slope of a line perpendicular to this line?
shtirl [24]

Answer:

there's no picture

Step-by-step explanation:

7 0
2 years ago
Consider the function f ( x ) = − 2 x 3 + 27 x 2 − 108 x + 9 . For this function there are three important open intervals: ( − [
Darya [45]

Answer:

<em>A=3 and B=6</em>

Step-by-step explanation:

<u>Increasing and Decreasing Intervals of Functions</u>

Given f(x) as a real function and f'(x) its first derivative.

If f'(a)>0 the function is increasing in x=a

If f'(a)<0 the function is decreasing in x=a

If f'(a)=0 the function has a critical point in x=a

As we can see, the critical points may define open intervals where the function has different behaviors.

We have

f ( x ) = - 2 x^3 + 27 x^2 - 108 x + 9

Computing the first derivative:

f' ( x ) = - 6 x^3 + 54 x - 108

We find the critical points equating f'(x) to zero

- 6 x^3 + 54 x - 108=0

Simplifying by -6

x^2 -9 x +18=0

We get the critical points

x=3,\ x=6

They define the following intervals

(-\infty,3),\ (3,6),\ (6,+\infty)

Thus A=3 and B=6

3 0
3 years ago
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