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3 years ago

10

I need help on this thank you

2 answers:

melamori03 [73]3 years ago

6 0

Answer:

b) BE ll CG

because the angle E and G are equal

zaharov [31]3 years ago

5 0

Answer:

B. lines BE and CG have to be parallel.

Step-by-step explanation:

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A block of wood measures 9.5 inches by 1.5 inches by 8 inches. What is the volume of the block of wood?

MatroZZZ [7]

114 is the answer hope this helps good luck and god bless!

6 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

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

7 0

2 years ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s). Points A and B

8090 [49]

Answer:

answer is 11 12

Step-by-step explanation:

4 0

4 years ago

In ΔPQR, the measure of ∠R=90°, the measure of ∠Q=46°, and RP = 97 feet. Find the length of PQ to the nearest tenth of a foot.

Whitepunk [10]

Answer:

PQ = 134.9 feet

Step-by-step explanation:

sin 46 = 97/PQ

0.7193 = 97/PQ

PQ = 134.85 feet

5 0

3 years ago

Svetlanka [38]

Answer:

Option B is the right choice.

Step-by-step explanation:

Given:

An are CE and an inscribed angle CBE.

Measure of inscribed angle CBE = 25 °

We have to find the measure of the arc CE.

Concept:

An inscribed angle is an angle with its vertex on the circle.
The measure of an inscribed angle is half the measure the intercepted arc.
Measure of inscribed angle = 1/2 × measure of intercepted arc .

To find the measure of arc CE.

⇒ $\angle CBE = \frac{1}{2} \times m\ (arc\ CE)$

⇒ $2\times \angle CBE = m\ (arc\ CE)$

⇒ $2\times 25 = m\ (arc\ CE)$

⇒ $50 = m\ (arc\ CE)$

Measure of intercepted arc CE = 50 degrees.

The measure of arc CE = 50° so, option B is the right choice.

4 0

3 years ago