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natulia [17]
2 years ago
10

A cylinder has a base diameter of 8 inches and a height of 7 inches. What is its volume in cubic inches, to the nearest tenths p

lace?
Mathematics
1 answer:
DaniilM [7]2 years ago
7 0

Answer:

351.9 cubic in

Step-by-step explanation:

The volume of a cylinder is hr²π.

r is half the diameter, so 8/2 is 4.

Plug in your h and r and get:

7 * 4² * π

7 * 16 * π

112π

Multiply the π and you'll get about 351.9 cubic in

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Plz help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
iren2701 [21]

Answer:

88 inches

Step-by-step explanation:

The given bicycle has a tyre that is 28 inches in diameter.

How far the bicycle moves forward each time the wheel goes around is the circumference of the bicycle tyre.

This is calculated using the formula:

C =\pi \: d

We substitute the diameter and

\pi =  \frac{22}{7}

C =  \frac{22}{7}  \times 28

This simplifies to

C =22 \times 4

88 \: inches

7 0
3 years ago
Can an absolute value equal a negative?
Montano1993 [528]
The distance of it away from zero cannot be a negative distance, distance is always positive therefor the absolute value will always be positive.

Short answer: no, absolute value is always positive

I hope this helps :)
7 0
2 years ago
Is 9x=270 addition property of equality or symmetric property?
Ede4ka [16]
This is the symmetric property, for you divide 9 to both sides to get x, instead of adding (addition property of equality)

So symmetric property should be your answer
Hope this helps
7 0
3 years ago
Read 2 more answers
Iv)<br>6x+3y=6xy<br>2x + 4y= 5xy​
Margaret [11]

Answer:

Ok, we have a system of equations:

6*x + 3*y = 6*x*y

2*x + 4*y = 5*x*y

First, we want to isolate one of the variables,

As we have almost the same expression (x*y) in the right side of both equations, we can see the quotient between the two equations:

(6*x + 3*y)/(2*x + 4*y) = 6/5

now we isolate one off the variables:

6*x + 3*y = (6/5)*(2*x + 4*y) =  (12/5)*x + (24/5)*y

x*(6 - 12/5) = y*(24/5  - 3)

x = y*(24/5 - 3)/(6 - 12/5) = 0.5*y

Now we can replace it in the first equation:

6*x + 3*y = 6*x*y

6*(0.5*y) + 3*y = 6*(0.5*y)*y

3*y + 3*y = 3*y^2

3*y^2 - 6*y = 0

Now we can find the solutions of that quadratic equation as:

y = \frac{6 +- \sqrt{(-6)^2 - 4*3*0} }{2*3} = \frac{6 +- 6}{6}

So we have two solutions

y = 0

y = 2.

Suppose that we select the solution y = 0

Then, using one of the equations we can find the value of x:

2*x + 4*0 = 5*x*0

2*x = 0

x = 0

(0, 0) is a solution

if we select the other solution, y = 2.

2*x + 4*2 = 5*x*2

2*x + 8 = 10*x

8 = (10 - 2)*x = 8x

x = 1.

(1, 2) is other solution

8 0
2 years ago
Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?
cluponka [151]

Under the given transformation, the Jacobian and its determinant are

\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2

so that

\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv

where D' is the region D transformed into the u-v plane. The remaining integral is the twice the area of D'.

Now, the integral over D is

\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y

but through the given transformation, the boundary of D' is the set of equations,

\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}

We require that u>0, and the last equation tells us that we would also need v>0. This means 1\le v\le\sqrt2 and 0, so that the integral over D' is

\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6

4 0
2 years ago
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