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2 years ago

If two sides are 36 feet what length is the long side and what type of triangle is it?

Mathematics

1 answer:

Aleks04 [339]2 years ago

5 0

Answer: maybe 50.9

Step-by-step explanation: we can use the pythagorean theorem A^+B^=C^

36 x36 = 1,296

1,296 + 1,296 = 2,592

square root of 2,592 = 50.911688245431421756860794071549 rounded to 50.9

so your answer maybe 50.9

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It snowed for 6 days in November

Step-by-step explanation:

Since there are 30 days in November

and 20 goes into 100 five times,

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2 years ago

Points A and B lie on a circle centered at point O. If what is the ratio of the area of sector AOB to the area of the circle?

Nataly [62]

Answer:

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Step-by-step explanation:

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2 years ago

What is X>195 on a number line

zlopas [31]

Step-by-step explanation:

anything greater than 195.

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2 years ago

Read 2 more answers

Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. One of the x-interc

Marina CMI [18]

3x^2 + 6x - 10 = 0
3(x^2 + 2x) - 10 = 0
3[ (x + 1)^2 - 1 ] - 10 = 0
3(x+1)^2 - 13 = 0

so the vertex is at (-1,-13)
the roots will be same distance from x = -1
that is a distance 1.08 --1 = 2.08

so other root is approximately -1 -2.08 = -3.08

the other intercept is at (-3.08,0)

5 0

3 years ago

Read 2 more answers

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

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

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3 years ago