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3 years ago

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What is the solution to the equation StartFraction 1 Over h minus 5 EndFraction + StartFraction 2 Over h + 5 EndFraction = Start

Fraction 16 Over h squared minus 25 EndFraction?
h = eleven-thirds
h = 5
h = 7
h = twenty-one-halves

2 answers:

vlabodo [156]3 years ago

7 0

Answer:

h=7

Step-by-step explanation:

djyliett [7]3 years ago

4 0

Answer:

h=7

Step-by-step explanation:

We want to solve the equation:

$\frac{1}{h-5}+\frac{2}{h+5}=\frac{16}{h^2-25}$

We multiply through by the LCM: $h^2-25=(h+5)(h-5)$

$(h^2-25*\frac{1}{h-5}+(h^2-25)*\frac{2}{h+5}=\frac{16}{h^2-25} \times(h^2-25)$

Simplify to get:

$h+5+2(h-5)=16$

We expand to get:

$h+5+2h-10=16$

$h+2h=16+10-5$ $3h=21$

h=7

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Calculus Problem

Roman55 [17]

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

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

7 0

2 years ago

THE NUMBER 1-10 ARE IN A BAG, YOU TAKE 2 OUT WITHOUT REPLACEMENT. WHAT IS THE PROBABILITY OF DRAWING THE NUMBER 7 THE FIRST TIME

ehidna [41]

Answer:

1 / 18

Step-by-step explanation:

PROBABILITY OF DRAWING THE NUMBER 7 THE FIRST TIME = Number of favorable outcome / Total number of outcomes

Favorable outcome = 1

Total number of outcome (1,2,3,4,5,6,7,8,9,10) = 10

PROBABILITY OF DRAWING THE NUMBER 7 THE FIRST TIME = 1 / 10

PROBABILITY OF DRAWING AN EVEN NUMBER THE SECOND TIME = Number of favorable outcome / Total number of outcomes

Favorable outcome (2,4,6,8,10) = 5

Total number of outcomes (1,2,3,4,5,6,8,9,10) = 9

PROBABILITY OF DRAWING AN EVEN NUMBER THE SECOND TIME = 5/9

PROBABILITY OF DRAWING THE NUMBER 7 THE FIRST TIME AND THEN DRAWING AN EVEN NUMBER = 1 / 10 * 5 / 9 = 1 / 18

6 0

3 years ago

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sattari [20]

I think it is 57 I just subtracted the two numbers

4 0

3 years ago

A cone has a volume of 226.08 cubic millimeters and a height of 6 millimeters. What is its

dezoksy [38]

Answer:

r≈5.36656314

Step-by-step explanation:

Since A=πr(r+ $\sqrt{h^2+r^2}$ ) so we get that 226.08/pi≈72 or 226.08/3.14=72 we then plug it in to get 72=r(r+ $\sqrt{36+r^2}$ ) so we get r≈5.36656314

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ELEN [110]

Answer:

They are parallel

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