Write an equivalent expression.

Mathematics

1 answer:

Stella [2.4K]3 years ago

8 0

7•7•7
2•2•2

7^3=343
2^3= 8
343•8
=2744

(7•7•7)x(2•2•2)

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Solve by any method from this unit. y2 - 5y = 3 Please help

ExtremeBDS [4]

Answer:

-1

Step-by-step explanation:

First we combine like terms, so y2 - 5y = - 3y

Now we get - 3y = 3

Then we divide both sides by - 3

Now we get y =-1

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Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.

Romashka [77]

By definition, the average change of rate is given by:
$AVR = \frac{f(x2)-f(x1)}{x2-x1}$
We will calculate AVR for each of the functions.
We have then:

f(x) = x^2 + 3x interval: [-2, 3]:
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f(3) = x^2 + 3x = (3)^2 + 3(3) = 9 + 9 = 18$
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$AVR = \frac{-20}{-5}$
$AVR = 4$

f(x) = 3x - 8 interval: [4, 5]:
$f(4) = 3(4) - 8 = 12 - 8 = 4 f(5) = 3(5) - 8 = 15 - 8 = 7$
$AVR = \frac{7-4}{5-4}$
$AVR = \frac{3}{1}$
$AVR = 3$

f(x) = x^2 - 2x interval: [-3, 4]
$f(-3) = (-3)^2 - 2(-3) = 9 + 6 = 15

f(4) = (4)^2 - 2(4) = 16 - 8 = 8$
$AVR = \frac{8-15}{4+3}$
$AVR = \frac{-7}{7}$
$AVR = -1$

f(x) = x^2 - 5 interval: [-1, 1]
$f(-1) = (-1)^2 - 5 = 1 - 5 = -4

f(1) = (1)^2 - 5 = 1 - 5 = -4$
$AVR = \frac{-4+4}{1+1}$
$AVR = \frac{0}{2}$
$AVR = 0$

Answer:
these functions from the greatest to the least value based on the average rate of change are:
f(x) = x^2 + 3x
f(x) = 3x - 8
f(x) = x^2 - 5
f(x) = x^2 - 2x

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3 years ago

Which statement about is true? –3 is an extraneous solution because solutions can never be negative. –3 is an extraneous solutio

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Answer:

Step-by-step explanation:

–3 is a valid solution because substituting it into the original equation gives a true statement.

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