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3 years ago

There are rational numbers that are integers / true or false

Mathematics

1 answer:

nikitadnepr [17]3 years ago

6 0

This is a true statement.

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What is the equation of a circle with center (8, 6) and radiuss5?

balu736 [363]

Answer:

Step-by-step explanation:

(8,6,4)

8 0

3 years ago

Is y=10 a function?

rewona [7]

Answer:

Yes

Step-by-step explanation:

x=10 would not however since it is a vertical line and fails the vertical line test where an x value cannot be repeated more than once.

y=10 would be a horizontal line coming from the point (0,10)

7 0

3 years ago

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All the dimensions of a parallelogram were multiplied by the same factor. If the area of the parallelogram went from 3 square ce

postnew [5]

Answer:

I was multiplied by 4.

Step-by-step explanation:

Because 3 times 1 is not 12

and 3 times 2 is not 12

and 3 times 3 is not 12

but only 3 time 4 is 12.

3 0

3 years ago

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Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago

Twice c is greater than 28

OLEGan [10]

I think C would equal 15, because 14 + 14 = 28; 15 + 15 = 30 which is greater than 28.

6 0

2 years ago