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kompoz [17]
3 years ago
12

a rectangular prism is 11 3/5 meters long ,9 meters wide ,and 12 1/2 meters high. what is the volume,in cubic meters,of the pris

m
Mathematics
1 answer:
Katen [24]3 years ago
4 0
The volume of this prism is 
V=l*w*h
rewrite the mixed number as fractions
V= 11 3/5* 9* 12 1/2=
V= 58/5 *9/1* 24/2
V=58*9*24/5*2
V=12528/10=1252 8/10 m^3
or
V=1252.80m^3

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If line segment RU is considered the base of parallelogram RSTU, what is the corresponding height of the parallelogram?
klasskru [66]
Given that line segment RU with vertices R(1, 1) and U(4, 5) is considered the base of parallelogram RSTU.

Then, the line segment ST with vertices S(7, 0) and T(10, 4) is the top of the parallelogram.

The corresponding height of the parallelogram is the length of a line with endponts at RU and ST and perpendicular to both RU and ST.

The equation of the line segment RU is given by
\frac{y-1}{x-1} = \frac{5-1}{4-1} = \frac{4}{3}  \\  \\ 3(y-1)=4(x-1) \\  \\ 3y-3=4x-4 \\  \\ 3y=4x-1 \\  \\ y= \frac{4}{3} x- \frac{1}{3}

Recall that given that two lines are perpendicular, the product of the slope of the two lines is -1.
Let the slope of the line perpendicular to line RU be m, then
\frac{4}{3} m=-1 \\  \\ m=- \frac{3}{4}

Thus, the equation of the line perpendicular to RU passing through point (1, 1) is given by
y-1=- \frac{3}{4} (x-1) \\  \\ 4(y-1)=-3(x-1) \\  \\ 4y-4=-3x+3 \\  \\ 4y=-3x+7 \\  \\ y=- \frac{3}{4} x+ \frac{7}{4}

The equation of the line segment ST is given by
\frac{y-0}{x-7} = \frac{4-0}{10-7} = \frac{4}{3}  \\  \\ 3y=4(x-7)=4x-28 \\  \\ y= \frac{4}{3} x- \frac{28}{3}

The line perpendicular to line segment RU intersected line segment ST at the point given by
- \frac{3}{4} x+ \frac{7}{4}=\frac{4}{3} x- \frac{28}{3} \\  \\ \frac{4}{3} x+\frac{3}{4} x=\frac{7}{4}+\frac{28}{3} \\  \\  \frac{25}{12} x= \frac{133}{12}  \\  \\ x= \frac{133}{25}  \\  \\ y=\frac{4}{3} \left(\frac{133}{25}\right)- \frac{28}{3}= -\frac{56}{25}

Thus, the corresponding height of the parallelogram is the line with endpoints
(1,1) \ and \ \left(\frac{133}{25},-\frac{56}{25}\right)

Recall that the length of a line passing through points
(x_1,y_1) \ and \ (x_2,y_2)
is given by
l= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Thus, the length of the line passing through points
(1,1) \ and \ \left(\frac{133}{25},-\frac{56}{25}\right)
is given by
l= \sqrt{\left(\frac{133}{25}-1\right)^2+\left(-\frac{56}{25}-1\right)^2}  \\  \\ = \sqrt{\left( \frac{108}{25}\right)^2+\left(- \frac{81}{25} \right)^2}= \sqrt{ \frac{11,664}{625} + \frac{6,561}{625} }  \\  \\ = \sqrt{ \frac{729}{25} } = \frac{27}{5} =5.4

Therefore, <span>the corresponding height of the given parallelogram is 5.4 units</span>
8 0
4 years ago
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Vaselesa [24]

Answer:

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Step-by-step explanation:

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3 0
3 years ago
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The volume of a cylinder is given by

V = A_b\cdot h

where A_b is the base area, and h is the height.

We are already given the height, which is 10.

We can deduce the base area since we are given a diameter: the radius is half the diameter, so it's 8, and the area of a circle is given by

A_b = \pi r^2 = \pi 8^2 = 64\pi

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V = A_b\cdot h = 64\pi \cdot 10 = 640\pi \approx 2010.6193\ldots

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