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3 years ago

What is the value of x? x = [ ? 10

Mathematics

1 answer:

kompoz [17]3 years ago

4 0

9514 1404 393

Answer:

Step-by-step explanation:

The triangle inside the circle is an equilateral triangle, so the central angle is 60°. That angle is one of the acute angles in the right triangle formed by the tangent line. The other acute angle (x°) in that right triangle is the complement of 60°.

x° = 90° -60° = 30°

x = 30

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The figure is made up of two identical quarter circles and a rectangle. The length of the rectangle is 35 cm and its breadth is

Ymorist [56]

Area of the middle rectangular section = length x width = 35 x 20 = 700 square cm.

The radius of the circular sections is 25 cm

Area of a full circle is pi x r^2 = 3.14 x 25^2 = 1962.5 square cm.

There are two quarter circles which is 1/2 of a circle. The area of those would be 1962.5/2 = 981.25 square cm.

Total area of the figure = 700 + 981.25 = 1681.25 square cm

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2 years ago

Please draw and label two perpendicular line segments KL and MN

muminat

So draw a cross! One line segment is KL and the other Mn

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2 years ago

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Suppose that 62 percent of the graduates from your high school go on to four-year colleges, 15 percent go on to two-year college

inn [45]

Answer:

b) .474

Step-by-step explanation:

If 62% go to a four-year college, that means that those who don't represent 38% of the high-school graduates.

You pick up someone who is NOT going to a four-year college (so, he's among the 38%)... what's the chance he's in the 18% of the whole high-school graduates population that found a job?

To calculate that probability, we have to divide 18% by 38%.

P = 18% / 38% = 0.4736, so 0.474

Since we are sure he doesn't go to a four-year college, there's 47.4% of chances he finds a job.

3 0

3 years ago

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

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

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2 years ago

Who likes factoring polynomials?

Diano4ka-milaya [45]

Answer:

me cuz its a solid brain workout

Step-by-step explanation:

7 0

2 years ago