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4 years ago

What is 3x-2y=1 -----------9x-6y=3

1 answer:

7 0

Solving for the value that makes both true

dvide the left equation by 3 to simplify
3x-2y=1
oh, they are same equaiton

inifnite solutions
the solutions are the points on the line y=(3/2)x-(1/2)

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A box contains four red balls and eight black balls. Two balls are randomly chosen from the box, and are not replaced. Let event

Leviafan [203]

Answer:

P(B) = 8/12

P(R | B) = 4/11

P(B ∩ R) = 8/33

The probability that the first ball chosen is black and the second ball chosen is red is about 24 percent.

Step-by-step explanation:

I just got it right on edge 2020. Good luck with school!!

6 0

3 years ago

Read 2 more answers

What is the height of the cone if its volune is 8,138.88 feet CUBED? Use 3.14 for pi. (Radius of base of cone is 12 ft)

Morgarella [4.7K]

Answer:

h = 54 feet

Step-by-step explanation:

The equation for the volume of a cone is V = $\frac{1}{3}$ × π × r² × h.

You have the volume, which is 8,138.88, and the radius of the base, which is 12.

V = 8,138.88

h = 12

π = 3.14

When you plug everything you know into the equation, you get:

V = $\frac{1}{3}$ × 3.14 × r² × h.

8,138.88 = $\frac{1}{3}$ × 3.14 × (12)² × h

After you multiply everything on the right side, you come out with:

8,138.88 = 150.72 × h

To find h, you can divide each side by 150.72.

8,138.88 ÷ 150.72 = 54 feet!

3 0

3 years ago

Addmaths probability. I need helpp for question a and b thank youu

motikmotik

Answer:

sorry this is soo blur

please follow me and mark has brilliant please

7 0

3 years ago

To find the area of a triangle, you can use the expression b × h ÷ 2, where b is the base of the triangle and h is its height. W

Zolol [24]

Answer:

Step-by-step explanation:

b x h is finding a square or rectangle, but since a triangle is half of a rectangle or square you can divide that by 2. So, 8 (b) x 4 (h) = 32, 32 / 2 = 16. Hope this helps! :D

6 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