All categories

2 years ago

How do you solve for p? -2p +14 > 4

Mathematics

1 answer:

IceJOKER [234]2 years ago

7 0

Answer:

subtract 14 from 4..then divide it by -2

your answer is then p>5

Step-by-step explanation:

hope this helpssss:)

You might be interested in

Please help solve this

marta [7]

Answer:

x = $\frac{5}{12}$

8 0

3 years ago

Read 2 more answers

What is the mean of the data set?

svetoff [14.1K]

B) 12.75
Just add them all together
10+15+14+8+18+11+12+12+10+10+17+16=153
The devide the anser by the ammount of numbers you added together
153÷12=12.75

8 0

2 years ago

Read 2 more answers

17 + x =31 <br> i need help

Gelneren [198K]

Answer:

x= 14

Step-by-step explanation:

First (and only) we subtract 17 from both sides :

17 + x - 17 = 31 - 17

x = 31 - 17

x = 14

Hope this helped and have a good day

4 0

2 years ago

Read 2 more answers

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 please please

monitta

Answer:

x=10

y=-2

(10,-2)

Step-by-step explanation:

If X=10, we can put 10 in for x in the equation.

30 + 5y = 20

Next, we need to solve to find y.

We will get y on its own so we subtract 30 from 20 to get

5y=-10

We then know that 5 goes into -10 -2 times.

y=-2

5 0

3 years ago