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3 years ago

Show how 7000÷1000 is the same as 7000÷10÷10÷10

Mathematics

2 answers:

drek231 [11]3 years ago

8 0

Because it has 3 zeros

pentagon [3]3 years ago

4 0

Answer:

it is because 10 times 100 is 1000 so it would be the same thing

Step-by-step explanation:

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How did he get from 3^-16/3/3^-8/3 to 0.00285333857/0.00285333857??? (The circled part) Because when I put it in the calculator

seropon [69]

Step-by-step explanation:

$\frac{ {3}^{ - \frac{16}{ 3} } }{ {3}^{ - \frac{ 8}{3} } } = \frac{0.002853338}{0.05341665075}$

I think, he made a mistake because the exponents have different values but then if you'll look at his results, he just wrote the same thing. If you want, you could post the original problem, and I'll solve it for you :)

7 0

3 years ago

What is the value of x?

zaharov [31]

Answer: around 47-49

Step-by-step explanation:

5 0

3 years ago

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Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

3 0

3 years ago

Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 7y = sin(x)

Elza [17]

The first thing we must do in this case is find the derivatives:
y = a sin (x) + b cos (x)
y '= a cos (x) - b sin (x)
y '' = -a sin (x) - b cos (x)
Substituting the values:
(-a sin (x) - b cos (x)) + (a cos (x) - b sin (x)) - 7 (a sin (x) + b cos (x)) = sin (x)
We rewrite:
(-a sin (x) - b cos (x)) + (a cos (x) - b sin (x)) - 7 (a sin (x) + b cos (x)) = sin (x)
sin (x) * (- a-b-7a) + cos (x) * (- b + a-7b) = sin (x)
sin (x) * (- b-8a) + cos (x) * (a-8b) = sin (x)
From here we get the system:
-b-8a = 1
a-8b = 0
Whose solution is:
a = -8 / 65
b = -1 / 65
Answer:
constants a and b are:
a = -8 / 65
b = -1 / 65

4 0

3 years ago

Find the simple interest you are investing $200 with a 5.2% interest rate of 3 years

kipiarov [429]

Answer:

Step-by-step explanation:

Simple interest is

I=Prt, where I= interest, P=principle (initial value), r=interest rate, t=time

I=200(0.052)3

I=$31.20

5 0

3 years ago

Read 2 more answers