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4 years ago

13

Express with positive exponents

1 answer:

Svetlanka [38]4 years ago

5 0

Answer:

2b^2

--------------

5a^4

Step-by-step explanation:

2a^-4

-------------

5b^-2

Negative exponents go to the denominator when they are in the numerator

and to the numerator when they are in the denominator

a^-4 becomes 1/a^4

1/b^-2 becomes b^2

2/5 * 1/a^4 * b^2

2*b^2 * 1/5a^4

Rewriting

2b^2

--------------

5a^4

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Which of these is most likely a pair of independent variables? a person’s income and savings the price of steel and the price of

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Independent variables are things that can stand alone. Which set of these variables does not rely on each other? For example the income and savings. I want 1,000,000 in my savings account, however my income is not nearly that high since I only make 19,000. Look at the other pairs and see if there is a correlation between them.

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4 years ago

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56.27 in expanded form

Lana71 [14]

Answer:

Step-by-step explanation:

56.27 =

50

+ 6

+ 0.2

+ 0.07

Expanded Factors Form:

56.27 =

5 × 10

+ 6 × 1

+ 2 × 0.1

+ 7 × 0.01

Expanded Exponential Form:

56.27 =

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Word Form:

56.27 =

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3 years ago

A jeweler wants to minimize business costs and has found that the average cost in dollars per necklace is given by C(x)=0.5x^2-7

Vika [28.1K]

Answer:

The maximum cost is 41 dollars per necklace.

Step-by-step explanation:

The average cost in dollars/necklace is given by :

$C(x)=0.5x^2-7x+65.5$ .....(1)

Where

x is the number of necklaces that are created

To maximize the cost, find the value of x by putting dC/dx = 0

$\dfrac{d}{dx}(0.5x^2-7x+65.5)=0\\\\x-7=0\\\\x=7$

Now put the value of x = 7 in equation (1).

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Hence, the maximum cost is 41 dollars per necklace.

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klemol [59]

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Find the y-intercept for the parabola defined by the equation y=x^2-x-12

o-na [289]

Answer:

The y-intercept is (0,-12).

Step-by-step explanation:

Given equation is $y=x^2-x-12$ .

Now question says to find the y-intercept of the given parabola $y=x^2-x-12$ .

We know that y-intercept means the y-value when x-value equals 0

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