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DanielleElmas [232]
3 years ago
8

X-1/8=5/24A . 1/12B. 5/24C. 1/3D. 1 2/3

Mathematics
1 answer:
mezya [45]3 years ago
4 0

Answer:

The answer c

Hope I helped :)

Step-by-step explanation:

Solve for x by simplifying both sides of the equation then isolating the variable.

Exact Form:

x=1/3

Decimal Form:

x=0.333333333

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The volume of a cone is 471.24 cubic inches and the height is 8 inches. What is the diameter
Vladimir [108]
The volume of a cone is V = 1/3 pi r^2 h
h = 8
pi = 3.14
V =471.24
r = ??? 
d = ???

471.24 = 3.14 * 8 * r^2
471.24 = 25.12 * r^2
471.24 / 25.12 = r^2
18.76 = r^2
r = sqrt(18.76)
r = 4.312

The diameter is 2*4.312 = 4.624 in. <<<<<==== answer.
7 0
4 years ago
Find the Volume of a triangular pyramid! ~
geniusboy [140]

Answer:

20

Step-by-step explanation:

using the given formula:

V = 1/3(Base Area)(height)

V = 1/3 (5)(12)

V = 1/3 (60)

V = 20

3 0
3 years ago
Read 2 more answers
Help any one please?
AfilCa [17]
A = (1/2)(8yd)(15yd) = 60 yd²
5 0
3 years ago
Solve the initial value problems.
slavikrds [6]

Both equations are linear, so I'll use the integrating factor method.

The first ODE

xy' + (x+1)y = 0 \implies y' + \dfrac{x+1}x y = 0

has integrating factor

\exp\left(\displaystyle \int\frac{x+1}x \, dx\right) =\exp\left(x+\ln(x)\right) = xe^x

In the original equation, multiply both sides by eˣ :

xe^x y' + (x+1) e^x y = 0

Observe that

d/dx [xeˣ] = eˣ + xeˣ = (x + 1) eˣ

so that the left side is the derivative of a product, namely

\left(xe^xy\right)' = 0

Integrate both sides with respect to x :

\displaystyle \int \left(xe^xy\right)' \, dx = \int 0 \, dx

xe^xy = C

Solve for y :

y = \dfrac{C}{xe^x}

Use the given initial condition to solve for C. When x = 1, y = 2, so

2 = \dfrac{C}{1\cdot e^1} \implies C = 2e

Then the particular solution is

\boxed{y = \dfrac{2e}{xe^x} = \dfrac{2e^{1-x}}x}

The second ODE

(1+x^2)y' - 2xy = 0 \implies y' - \dfrac{2x}{1+x^2} y = 0

has integrating factor

\exp\left(\displaystyle \int -\frac{2x}{1+x^2} \, dx\right) = \exp\left(-\ln(1+x^2)\right) = \dfrac1{1+x^2}

Multiply both sides of the equation by 1/(1 + x²) :

\dfrac1{1+x^2} y' - \dfrac{2x}{(1+x^2)^2} y = 0

and observe that

d/dx[1/(1 + x²)] = -2x/(1 + x²)²

Then

\left(\dfrac1{1+x^2}y\right)' = 0

\dfrac1{1+x^2}y = C

y = C(1 + x^2)

When x = 0, y = 3, so

3 = C(1+0^2) \implies C=3

\implies \boxed{y = 3(1 + x^2) = 3 + 3x^2}

7 0
2 years ago
Factor 9b^2 - 4.<br><br> (3b + 1)(3b - 4)<br> (3b + 2)(3b - 2)<br> (3b - 2)(3b - 2)
maria [59]

The answer should be (3b + 2)(3b - 2) since if you multiply those two, you will get back to original equation.

Hope this helps!

5 0
3 years ago
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