3 years ago

I need help on this please

Mathematics

1 answer:

irakobra [83]3 years ago

4 0

The answer is C :$396.00

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A.140 B. 60 C.120 or D.40 ?

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Answer:

A-140

Step-by-step explanation:

A starts at 0 and from the side you're measuring it would be 140 and not 40.

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Tammy is creating potpourri bowls using 40 bags of shredded bark and 20 bags of flower

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Step-by-step explanation:

She only has 20 bags of flower petals, so she cannot create more than that. Because 40 is divisible by 20, she can make 20 bowls.

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3 years ago

Which is not true about the random sample method?

USPshnik [31]

B. The sample is likely to represent one portion of the population.

This is impossible to say because it's random.

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4 years ago

How do you find slope of a line with rise and run? <br><br> Rise: 5<br> Run:2

eduard

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3 years ago

Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2

NNADVOKAT [17]

$y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}$

Divide both sides by $\dfrac13y^{-2}(x)$ :

$3y^2y'-y^3=xe^x\ln x$

Substitute $v(x)=y(x)^3$ , so that $v'(x)=3y(x)^2y'(x)$ .

$v'-v=xe^x\ln x$

Multiply both sides by $e^{-x}$ :

$e^{-x}v'-e^{-x}v=x\ln x$

The left side can be condensed into the derivative of a product.

$(e^{-x}v)'=x\ln x$

Integrate both sides to get

$e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C$

Solve for $v(x)$ :

$v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

Solve for $y(x)$ :

$y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

$\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}$

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3 years ago