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3 years ago

Consider the division problem 2^5 describe a situation this division could represent

1 answer:

4 0

That is not a division problem that is a 2 with an exponent of 5. But if you are talking about 2/5 here is a situation:
Five pies are split amongst two friends.

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The quotient of 2 numbers is -3 and their difference is -16. What are the 2 numbers?

n200080 [17]

Answer:

-24 / 8

= -3

8 - 24

= -16

so the 2 numbers are 8 and - 24

3 0

4 years ago

When verifying that a function represented in another table is the inverse of the given cost function, which ordered pair can be

babunello [35]

Answer:

Option B is correct.

Step-by-step explanation:

To find the inverse of the function, the ordered pairs are inverted. i.e. x becomes y and y becomes x.

So, We are given in the table:

(20,52) ,(25,67.5), (30,84)

The inverse would be:

(52,20), (67.5,25),(84,30)

So, Option B is correct.

5 0

3 years ago

Read 2 more answers

What does the constant of variation mean?

Ostrovityanka [42]

Answer:

The constant of variation is the number that relates two variables that are directly proportional or inversely proportional to one another.

Step-by-step explanation:

3 0

3 years ago

HELP ASAP photo attached

ankoles [38]

Answer:

D. undefined

General Formulas and Concepts:

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Trig Derivative: $\displaystyle \frac{d}{dx}[sinu] = u'cosu$

Derivatives of Parametrics: $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Step-by-step explanation:

Step 1: Define

$\displaystyle \frac{dx}{dt} = 5$

$\displaystyle \frac{dy}{dt} = sin(t^2)$

Step 2: Differentiate

[x Derivative] Basic Power Rule: $\displaystyle \frac{d^2x}{dt^2} = 0$
[y Derivative] Trig Derivative [Chain Rule]: $\displaystyle \frac{d^2y}{dt^2} = cos(t^2) \cdot \frac{d}{dt}[t^2]$
[y Derivative] Basic Power Rule: $\displaystyle \frac{d^2y}{dt^2} = cos(t^2) \cdot 2t^{2 - 1}$
[y Derivative] Simplify: $\displaystyle \frac{d^2y}{dt^2} = 2tcos(t^2)$
[Derivative] Rewrite: $\displaystyle \frac{d^2y}{dx^2} = \frac{2tcos(t^2)}{0}$