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3 years ago

How do I take 8 times 5/12

Mathematics

1 answer:

Butoxors [25]3 years ago

6 0

U put 8/1 times 5/12 and multiply straight across and get 40/12

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The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original

Zanzabum

Answer:

The original number could be 85.

Step-by-step explanation:

Let the 2 digits be x and y.

Let the number be xy then, assuming that x is the larger digit:

x - y = 3.

x = y + 3

Also

10y + x + 10x + y = 143

Substituting for x:

10y + y + 3 + 10(y + 3) + y = 143

22y + 33 = 143

22y = 110

y = 5.

So x = y + 3 = 8.

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3 years ago

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True or false, the fractions ½ and ¾ are equivalent.<br><br> True<br><br> False

madreJ [45]

False

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2 years ago

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Best estimate for 61/7

Ne4ueva [31]

I would say the best estimate would be 9 because the real answer is 8.71 and if you round that to the 1 place the answer would be 9.

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2 years ago

Which angles are linear pairs. check all that apply

snow_tiger [21]

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2 years ago

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Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

5 0

2 years ago