The weight of 4 eggs is shown. Identify the constant of proportionality of total weight to number of

In a marble collection, 1/8 of the marbles are blue. Of the blue marbles, 1/2 have sparkles. What fraction of the marbles in the

Delvig [45]

Answer:

1/16

Step-by-step explanation:

if half the amount of blue marbles have sparkles you need to divide the amount of blue marbles in half

so 1/2 of 1/8 is 1/16

8 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

Please help and show work I’m begging

mixas84 [53]

13. Pick a point and see which formula works.
Ay = -4, A'y = 7. Only the formula of selection D makes that translation.

14. Use the compound interest formula A = P*(1 +r/n)^(nt).
..1500*1.015^80 = 4935.99, matching selection C

15. The lid has a perimeter of 90", so the area of the sides is
.. 90" * 24" = 2160 in^2
The area of the lid is
.. 30" * 15" = 450 in^2

The gray area is (2160 -450) in^2 = 1710 in^2 larger, corresponding to selection C.

16. The only formula that maps (7, -1) to (21, -3) is that of selection D.

_____
The middle two problems are the only ones that require you to have prior knowledge. The others could be answered simply by seeing if the answers work.

5 0

3 years ago

Two cars start at the same point. One travels north at 65 km/h and the other travels east at 50 km/h. How far are they apart aft

Kisachek [45]

Answer:

they travel 115 km/h long

4 0

2 years ago

Can anybody help me?

lakkis [162]

You know that 6x+3 equals 45 because the two sides are equal. So you first put the equation 6x+3=45. Then you use the subtraction property of equality and subtract 3 from both sides. This gives you 6x=42. Then you divide 6 from both sides to get you final answer, x=7

8 0

3 years ago