All categories

3 years ago

How to check volume of cabinets 24by72by18

Mathematics

2 answers:

Pachacha [2.7K]3 years ago

4 0

31,04 is your answer just multiple it and your answers there

Hope this helped

mash [69]3 years ago

3 0

In order to find the volume of the cabinet, use the volume formula for a rectangular prism, which is the shape of the cabinet.

lwh Formula for volume of a rectangular prism
24 x 72 x 18 Now, plug in each of the values you've been given.
31,104 Multiply the numbers; you might want to use paper and pencil, or a calculator, if you're allowed to.

The volume of the cabinet is 31,104 un³.

Hope this helps!

You might be interested in

A test has 20 questions. If peter gets 80% correct, how many questions did peter missed?

Umnica [9.8K]

16 because if you put 16 into a percentage it's 80. 16 divide 20 times 100 equal 80

8 0

2 years ago

Read 2 more answers

What is the longest line segment that can be drawn in a right rectangular prism that is 15cm long, 13cm wide, and 8cm tall?

Effectus [21]

Answer:

15cm long

Step-by-step explanation:

8 0

2 years ago

Can someone please help me with this? ^^"

meriva

Answer:

Not issue discussed plz full page

7 0

2 years ago

Which is true from these equations

e-lub [12.9K]

Answer:

B and E

Step-by-step explanation:

4 0

2 years ago

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

Read 2 more answers