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3 years ago

Round 8,088 to the nearest thousand

Mathematics

2 answers:

rusak2 [61]3 years ago

6 0

It is 8,000 because it is isn't above 8,500

GalinKa [24]3 years ago

3 0

8,088 rounds to 8,000 to the nearest thousand.

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Q1 Write the number in expanded form for the significant digits ( as fractions)

Dimas [21]

To write in expanded form is to simply mean writing each number into its place value.

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1 year ago

Please help me with this

juin [17]

Answer:

10π (31.415926)

√101(10.0498756)

9.92749...

0.99853...

Step-by-step explanation:

For 10π, we can simplify π into 3.14 for an approximate amount. 3.14*10=31.4

For the square root of ten, we can estimate it's between 10 and 11 since 10 squared is 100 and 11 squared is 121.

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2 years ago

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What two products would you add to find 713x48

umka21 [38]

9514 1404 393

Answer:

40·713 and 8·713

Step-by-step explanation:

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2 years ago

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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

2 years ago

Summary

slava [35]

It the multiplication property because you are multiplying 2x=20 so then you divide on both sides with 2 and that gives you 10

5 0

2 years ago