All categories

2 years ago

Help please I really need help :)

Mathematics

1 answer:

Firdavs [7]2 years ago

6 0

Answer:

Solution given:

For 1st

x²=71

For x=± $\sqrt{71}$

and

For 2nd

x³=71

For x=± $\sqrt[3]{71}$

So answer is:

In first row

Solution to x²=71

In second row

Solution to x³=71

In third row

neither

In forth row

Solution to x³=71

You might be interested in

Why does totality during a lunar eclipse last longer than totality during a solar eclipse?

34kurt

Answer:

Step-by-step explanation:

Becuase the moon is completely covering the sun for 7 minutes

3 0

2 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

Solve for x x+0.15x=3.45

otez555 [7]

Answer: x should equal 3

5 0

3 years ago

Read 2 more answers

A birthday cake was cut into equal pieces, and four pieces were eaten. The fraction is 3/7 how many are left

GaryK [48]

3 are left.
3/7 means there were 7 slices, but it's missing 4 slices. (Eaten)

i hope this helps :)

7 0

3 years ago

Mr. Arnold drives 130 miles round trip to work 5 days a week. Write and solve an equation to find how many miles he drives in 60

mart [117]

Answer:

The answer is D.

Step-by-step explanation:

130 x 60 = 7,800 miles. you don't need the extra information "5 days a week" to solve the problem.

3 0

3 years ago