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3 years ago

Suppose f(x)=x^2 and g(x)=(1/3x)^2 which statement comepare the graph of f(x) to the graph of g(x)

Mathematics

1 answer:

statuscvo [17]3 years ago

8 0

Answer: c

Step-by-step explanation:

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Every even number,except 2,is composite

Degger [83]

Yes. That's correct. Is that what you're asking...? To clarify, prime numbers have a single factor (one and itself) , composite numbers have multiple factors.

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4 years ago

Is this all correct?

Doss [256]

Answer:

Step-by-step explanation:

11. Correct

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13. correct

13a. f(-1)=8

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3 years ago

A student says that a correlation coefficient of −0.93 indicates that the two quantities of a data set have a weak correlation.

zloy xaker [14]

Answer:

Step-by-step explanation:

Recall that correlation coefficients range from -1 to +1. So, -0.93 is a strong negative relationship. Two interpretations: 1) the correlation is actually very strong, and 2) the correlation is negative. When the correlation is negative, the dependent variable DECREASES when the independent variable INCREASES.

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3 years ago

Please help! I do not understand!

Sergio [31]

Your answer would be approximately 254.5 if you round since 9^2 times pie you would get such

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4 years ago

A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f

Andrews [41]

Answer:

$A = \frac{P}{r}\left( e^{rt} -1 \right)$

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

$\frac{dA}{rA+P} = dt$

Integration on both sides gives

$\int \frac{dA}{rA+P} = \int dt$

where $c$ is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

$\int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c$