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olga2289 [7]
3 years ago
7

00:00

Mathematics
1 answer:
Dafna1 [17]3 years ago
5 0

Answer:

B.

15% of 75

Step-by-step explanation:

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Please explain how to slove thank you.
Sergeeva-Olga [200]
The average rate of change of on the interval [a,b] is:

 f(b)−f(a)/ b-a
or
y1-y2/x1-x2 which is how I learned it, average ratw of change is just slope

so you would plug your numbers into x

1st the b: 6((1))^2+31((1))-12
=6+31-12 1^2=1, so it cancels
f(b) =25


2nd the a: 6((-2))^2+31((-2))-12
=6(4)+31(-2)-12
=24+-62-12
f(a) =-50

3rd plug them in: (25- -50)÷(1-(-2)
=75÷3
=25


so your average rate of change(or slope) is 25 :)

7 0
3 years ago
Solve question 5 all steps
valkas [14]

Answer:

Answer in the pic with steps

Hope it helps

6 0
2 years ago
What is the equation of the line whose y-Intercept is 3 and slope is 1?
Arada [10]

y = x + 3

as x increases by 1, y increases by 1 (Slope)

when x equals 0, y equals 3 (y-intercept)

5 0
3 years ago
A rectangular bin with an open top and volume of 38.72 cubic feet is to be built. The length of its base must be twice the width
KengaRu [80]

Answer:

Width=2.2 FT

Length=4.4 FT

Heigh= 4 FT

Step-by-step explanation:

5 0
3 years ago
Solving for matrices
muminat

Answer:

D

Step-by-step explanation:

The augmented matrix for the system of three equaitons is

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\5&2&-2&|&11\\5&-4&4&|&-7\end{array}\right)

Multiply the first row by 5, the second row by -3 and add these two rows:

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\5&-4&4&|&-7\end{array}\right)

Subtract the third row from the second:

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&6&-6&|&18\end{array}\right)

Divide the third row by 6:

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&1&-1&|&3\end{array}\right)

Now multiply  the third equation by 26 and add it to the second row:

\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&0&-45&|&-90\end{array}\right)

You get the system of three equations:

\left\{\begin{array}{r}3x-4y-5z=-27\\-26y-19z=-168\\-45z=-90\end{array}\right.

From the third equation

z=\dfrac{90}{45}=2.

Substitute z=2 into the second equation:

-26y-19\cdot 2=-168\\ \\-26y-38=-168\\ \\-26y=-168+38=-130\\ \\y=\dfrac{130}{26}=5.

Now substitute z=2 and y=5 into the first equation:

3x-4\cdot 5-5\cdot 2=-27\\ \\3x-20-10=-27\\ \\3x-30=-27\\ \\3x=-27+30=3\\ \\x=1.

The solution is (1,5,2)

4 0
3 years ago
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