I really need help on my engineering homework.

Write an equation in slope-intercept form for the line with slope 1/5 and y-intercept 5.

AVprozaik [17]

Answer:

y=1/5x+5

Step-by-step explanation:

Slope-intercept form is written y=mx+b.

y- needs to be by itself in this form.

m- the slope.

x- always to the right side of the slope, or m.

b- y-intercept.

Now that we know this, all we have to do is plug in the numbers we have. So when we plug this in, the equation looks like this:

y=1/5x+5

Hope this helps you and have a great day :3

8 0

3 years ago

Please help with the question below in the picture

erma4kov [3.2K]

Answer:

10.5 cm or 1.5 cm

Step-by-step explanation:

Case 1

O in between A and B

OA = 12 cm, OB = 9 cm

Distance between midpoints:

12/2 + 9/2 = 10.5 cm

Case 2

B in between O and A

OA = 12 cm, OB = 9 cm

Distance between midpoints:

12/2 - 9/2 = 1.5 cm

5 0

2 years ago

One third of the candidates for a job were 25 years old or younger. two sevenths of the candidates were at least 50 years old. i

Shalnov [3]

There are three groups of people: 25 years old and younger, between 25 to 50, and 50 years old and older. Their fractions must equal to 1 because these three together form the whole.

1 = 1/3 + 2/7 + x
where x is the fraction for people ages 25 to 50 years old
x = 8/21

So, the actual number of people ages 25 to 50 years old is:

84(8/21) = 32 people

7 0

3 years ago

Jason took his girlfriend to dinner at a nice restaurant. the bill was $84.20 and he needs to give a 15% tip. How much tip shoul

marysya [2.9K]

Answer:

Tip Total: $ 12.63

Tip Each Person: $ 6.32

Total (Bill + Tip): $ 96.83

Total Each Person: $ 48.42

Step-by-step explanation:

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