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3 years ago

Prism A is similar to Prism B. The volume of Prism A is 2080 cm³.

Mathematics

1 answer:

velikii [3]3 years ago

7 0

Given data:

Volume of the prism A = 2080 $c m^{3}$

Let us find the ratio of the volume of prisms given in the diagram:
$\frac{ V_{A}}{V_{B}} = \frac{8^{3}}{4^{3}}$
$\frac{ V_{A}}{V_{B}} = 8$

Since the condition is:
The volume of Prism A = The volume of Prism B
Therefore,
The ratio of their volumes should be 8.
$\frac{ V_{A}}{V_{B}} =8$

Since Volume of A = 2080 $cm^{3}$

Plug-in the value, you would get:

$\frac{2080}{8}} = V_{B}$

Ans: Volume of B = 260 $cm^{3}$

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1 year ago

Read 2 more answers

Twice a number plus twice a second number is 310. The difference between the numbers is 55

agasfer [191]

Answer:

x = 105

y = 50

Step-by-step explanation:

To solve this, try writing it out as an equation/expression first.

When you do, you should get something like this:

2x + 2y = 310

x - y = 55

This is a system of equations. To solve this, we're going to multiply the second equation [x - y = 55] by 2.

2(x - y = 55) = 2x - 2y = 110

Now, subtract your new equation from the first one [2x + 2y = 310]

2x + 2y = 310

-2x + 2y = -110

//If you're wondering why I changed the signs of the new equation, I multiplied our new equation [2x - 2y = 110] by -1 since we're subtracting//

When you're done subtracting, the "2x" and "-2x" cancel out, so you're left with:

4y = 200

Now, simply solve for "y" by dividing both sides by 4.

4y = 200

y = 50

Now that we know y's value, substitute it into one of our original equations to solve for x.

//I'm gonna plug y into the second equation [x - y = 55] just because it's easier, but you can plug it into the first one as well and get the same answer//

When we plug it in, we get:

x - (50) = 55

To solve for "x," we must get rid of "-50." To do this, add 50 to both sides.

x - 50 = 55

x = 105

So, x = 105 and y = 50

4 0

2 years ago

What percentage of the semi circle is shaded?

mezya [45]

The percentage of the semicircle shaded section is approximately 23,606 %.

The percentage of the area of the semicircle is equal to the ratio of the semicircle area minus the half-cross area to the semicircle area. In other words, we have the following expression:

$r = \left(\frac{A_{s}-A_{h}}{A_{s}} \right)\times 100\,\%$

$r = \left(1-\frac{A_{h}}{A_{s}} \right)\times 100\,\%$ (1)

Where:

$A_{h}$ - Area of the half cross, in square centimeters.
$A_{s}$ - Area of the semicircle, in square centimeters.
$r$ - Percentage of the shaded section of the semicircle.

And the percentage of the shaded section is:

$r = \left[1-\frac{4 \cdot (2\,cm)^{2}+4\cdot \left(\frac{1}{2} \right)\cdot (2\,cm)^{2}}{0.5\cdot \pi\cdot (16\,cm^{2}+4\,cm^{2})} \right]\times 100$