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3 years ago

You are given a cubic box with an edge length of 20in, what is the volume of the box?

Mathematics

1 answer:

larisa [96]3 years ago

5 0

Answer:

= 40 inches3 i think

Step-by-step explanation:

Vbox = l × w × h

Vbox = 5 inches × 4 inches × 2 inches

Vbox = 20 inches2 × 2 inches

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2) At the store Julie bought a shirt for $23.97, which was 25% off the original price. Which of these should she use to find x,

Scrat [10]

Answer:

0.75x = 23.97

Step-by-step explanation:

The total is 23.97. 0.25% is off so you have to subtracting 25% from 100% and you will get 75% if you convert it to decimal you will get 0.75.

5 0

2 years ago

If there are 5 boys and 7 girls, write the ratio of girls to total students

Marizza181 [45]

Answer:

7:12

Step-by-step explanation:

boys=5

girls=7

total=boys+girls

12=5+7

girls:total

7:12

3 0

2 years ago

Find the area ratio of a regular octahedron and a tetrahedron regular, knowing that the diagonal of the octahedron is equal to h

Anton [14]

Answer:

$\frac{4}{3}$

Step-by-step explanation:

The area of a regular octahedron is given by:

area = $2\sqrt{3}\ *edge^2$ . Let a is the length of the edge (diagonal).

area = $2\sqrt{3}\ *a^2$

Given that the diagonal of the octahedron is equal to height (h) of the tetrahedron i.e.

a = h, where h is the height of the tetrahedron and a is the diagonal of the octahedron. Let the edge of the tetrrahedron be e. To find the edge of the tetrahedron, we use:

$h=\sqrt{\frac{2}{3} } e\\but\ h=a\\a=\sqrt{\frac{2}{3} } e\\e=\sqrt{\frac{3}{2} }a$

The area of a tetrahedron is given by:

area = $\sqrt{3}\ *edge^2$ = $\sqrt{3} *(\sqrt{\frac{3}{2} }a)^2=\frac{3}{2}\sqrt{3} *a^2$

The ratio of area of regular octahedron to area tetrahedron regular is given as:

Ratio = $\frac{2\sqrt{3}\ *a^2}{\frac{3}{2} \sqrt{3}*a^2} =\frac{4}{3}$

7 0

3 years ago

Solve the right triangle. Round decimals to the nearest tenth

LekaFEV [45]

Answer: 42

Step-by-step explanation:

angle t is 42 degrees

5 0

1 year ago

I need help asap!!!!!!!!!!!!!

AnnZ [28]

Substituting x=6 and y=1 into both equations,

$x+2y=8 \\ \\ 6+2(1)=8 \\ \\ 8=8 \\ \\ \\ \\ 5x-2y=4 \\ \\ 5(6)-2(1)=4 \\ \\ 28=4$

Since the second equation is not satisfied, (6,1) is not a solution to the system of equations.

3 0

1 year ago

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