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2 years ago

Does anyone know a fraction equation that equals 13?

Mathematics

2 answers:

taurus [48]2 years ago

7 0

Answer:

11 1/2 + 1 1/2 = 13

Step-by-step explanation:

11+1=12

1/2+ 1/2 =1

12+1= 13

Angelina_Jolie [31]2 years ago

5 0

Answer:

There is a lot of fractions that can combine to a value of 13.

One can be 12/12 + 1200/100

A more complicated one might be 3 3/4 + 9 1/4

Step-by-step explanation:

Fractions are displayed like this : x/y

Mixed numbers are displayed like this : x x/y

12/12 + 1200/100 Simplify

1 + 12 Add

3 3/4 + 9 1/4 Add the wholes

12 + 3/4 + 1/4 Add the fractions

12 + 1 Add everything

<h3><u><em>Please rate this and please give brainliest. Thanks!!! </em></u></h3><h3><u><em>Appreciate it! : )</em></u></h3><h3 /><h2><u><em>And always, </em></u></h2><h2><u><em> SIMPLIFY BANANAS : )</em></u></h2>

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The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

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