What two tens does 192 fall between?

Candice plotted the points (2, 15) and (0, -1) and then drew a line through these two points. What is the slope of the line she

FrozenT [24]

Use the slope equation: $\frac{y2-y1}{x2-x1}$

Plug in the points (2, 15) and (0, -1).
$\frac{-1-15}{0-2}$ ⇒ $\frac{-16}{-2}$ = 8

The slope of the line is 8.

7 0

3 years ago

If Angela $98,760 home appreciates 3% a year will she have enough appreciation to try to sell a home for a $15,000 profit in fiv

olasank [31]

Answer:

Yes

Step-by-step explanation:

So, first, in 5 years, the home will have appreciated by 15%. (5 years times 3%). Once you find 15% of 98760, which is 658400, you have to add it on to the original price of the house. At this point, the house costs 757160 dollars. You then subtract the original price of the house from the price of the house 5 years from now. (757160-98760) and you get 658400. As you can tell, 658400>15000. Therefore, the answer is yes.

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

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

7 0

2 years ago

Find the value please help me

Lemur [1.5K]

The answer is 318

Hope this helps

3 0

3 years ago

Read 2 more answers

A box is known to hold exactly 125 in3. Which side dimensions produce the most accurate measurement of the box?

Harman [31]

D is the answer most likely

5 0

3 years ago

Read 2 more answers