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Arisa [49]
3 years ago
8

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 ?
GalinKa [24]

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.

5 0
3 years ago
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Tammy is creating potpourri bowls using 40 bags of shredded bark and 20 bags of flower
Anni [7]

Answer:

20

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.

6 0
4 years ago
How do you find slope of a line with rise and run? <br><br> Rise: 5<br> Run:2
eduard
\text{slope}=\dfrac{\text{rise}}{\text{run}}\\&#10;\text{slope}=\dfrac{5}{2}
3 0
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}}

4 0
3 years ago
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