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Veronika [31]
3 years ago
12

Cassie has a small cube shaped box. Its volume is 64 cubic cm. What is the area of one face of tje box.

Mathematics
1 answer:
Andrews [41]3 years ago
8 0

Answer:  One face is 16 square centimeters.

Step-by-step explanation: To find the area of a face, you need to know the length of the side. All the sides of the cube are the same, so take the cube root of 64.

\sqrt[3]{64}=4 cm is the length of the side

The area of one face is the side length squared 4² = 16

4cm × 4cm = 16 square centimeters.

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She puts the gifts into a box. She puts a label on one side of the box.How many faces do not have labels?faces.
Kisachek [45]
A closed box has 6 faces.

There is 1 label on 1 face.

Subtract 1 from 6

6 - 1 = 5

5 faces is your answer


Hope this helps
3 0
3 years ago
If (x^2 - y^2 = 10 and (x + y) =2 find x and y?
vampirchik [111]

Answer:

x = 3.5 ; y = -1.5

Step-by-step explanation:

5 0
3 years ago
Sally has three tiles each tile is has a different number on it so he puts the three tires down to make a number each number is
Juli2301 [7.4K]

this would be equal to factorial 3 which is written 3! .

It is 3 * 2 *  1 =  6 numbers answer.

4 0
4 years ago
The angles of a triangle are 2x, 3x, and 4x degrees. Find the value of x.
Wittaler [7]
**Remember this rule: the sum of a triangle's angle measurements always = 180
So 2x + 3x + 4x = 180
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A) 20
4 0
3 years ago
Please help me to prove this!<br>I need is no.(c). So, please help me do it.<br>​
zloy xaker [14]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = 90°                  → A + B = 90° - C

                                                     → C = 90° - (A + B)

Use the Double Angle Identity:      cos 2A = 1 - 2 sin² A

                                                       → sin² A = (1 - cos 2A)/2

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Product to Sum Identity: cos (A - B) - cos (A + B) = 2 sin A · sin B

Use the Cofunction Identities:      cos (90° - A) = sin A

                                                       sin (90° - A) = cos A

<u>Proof LHS → RHS:</u>

LHS:                       sin² A + sin² B + sin² C

\text{Double Angle:}\qquad \dfrac{1-\cos 2A}{2}+\dfrac{1-\cos 2B}{2}+\sin^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2-\cos 2A-\cos 2B\bigg)+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\sin^2 C

\text{Sum to Product:}\quad 1-\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\cos (A+B)\cdot \cos (A-B)+\sin^2 C

Given:                1 - cos (90° - C) · cos (A - B) + sin² C

Cofunction:       1 - sin C · cos (A - B) + sin² C

Factor:               1 - sin C [cos (A - B) + sin C]

Given:                1 - sin C[cos (A - B) - sin (90° - (A + B))]

Cofunction:       1 - sin C[cos (A - B) - cos (A + B)]

Sum to Product:       1 - sin C [2 sin A · sin B]

                            = 1 - 2 sin A · sin B · sin C

LHS = RHS: 1 - 2 sin A · sin B · sin C = 1 - 2 sin A · sin B · sin C   \checkmark

6 0
3 years ago
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