Find the area of the triangles with

A cube has a side length of 5 inches. What is the surface area, in square inches of the cube?

makkiz [27]

Answer:

A=6a2=6·52=150in²

Step-by-step explanation:

4 0

3 years ago

1.Which are the equation and solution for the sentence, "Anumber, n, divided by

Assoli18 [71]

Answer:

1) B/ 0.4n = 12; n = 30

2) B/ 125 – x = 58

3) B/ $1.75

Step-by-step explanation:

1) 0.4n = 12

n = 12 ÷ 0.4

n = 30

2) 125 – x = 58

– x = 58 – 125

– x = – 67 ( – & – will get cancelled)

x = 67

a) 125 + x = 58

x = 58 – 125

x = – 67 (pages won't be in negative)

c) 125 ÷ 58 = x

2.15 = x (wrong)

d) 58 – x = 125

– x = 125 – 58

– x = 67

x = 67 ÷ –1

x = – 67 ( pages won't be in negative)

3) 12x = 21

x = 21 ÷ 12

x = 1.75

3 0

3 years ago

Solve the equation y-12=4y

andre [41]

Hi there!

$y - 12=4y 

We need to subtract 4y from both sides 

y-12-4y = 4y - 4y 

3y - 12=0 

3y = 0 + 12 

3y=12 

Divide both sides by 3 

-3y/3 = 12/-3 

y= -4$

8 0

4 years ago

Yall this the last one to prove yourself

Alexandra [31]

3. y=112 this takes a lot of process of elimination so ill save you the paragraphs of numbers

second row - k³⁰/k¹⁴ - k¹⁶

third row - 12 ⁴⁵⁻²³ - 12²²

hope this helps

7 0

3 years ago

A steel safe with mass 2200 kg falls onto concrete. Just

Virty [35]

The kinetic energy of the safe increases the force exerted by the concrete

to several times the weight of the safe.

The magnitude of the force exerted on the safe by the concrete on the is approximately $\underline{29.\overline 3 \, \mathrm{MN}}$

The concrete exerts a force that is approximately 1,359.16 times the weight of the safe.

Reasons:

First part

The mass of the steel safe, m = 2,200 kg

Velocity of the safe just before it hits the concrete, v = 40 m/s

The amount by which the safe was compressed, d = 0.06 m

The kinetic energy, K.E., of the safe just before it hits the round is therefore;

$\displaystyle K.E. = \mathbf{\frac{1}{2} \cdot m \cdot v^2}$

$\displaystyle K.E._{safe} = \frac{1}{2} \times 2,200 \times 40^2 = 1,760,000 \ Joules$

Work done by concrete, W = Force, F × Distance, d

$\displaystyle Force, \, F = \mathbf{\frac{Work, \, W}{Distance, \, d}}$

By the law of conservation of energy, we have;

The work done by the concrete, W = The kinetic energy, K.E. given by the safe

W = K.E. = 1,760,000 J

The effect of the work = The change in the height of the safe

Therefore;

The distance, d, over which the force of the concrete is exerted = The change in the height of the safe = 0.06 m

d = 0.06 m

Therefore;

$\displaystyle The \ force \ of \ the \ concrete, \, F = \frac{1,760,000\, J}{0.06 \, m} = 29,333,333. \overline 3 \, N = 29.\overline 3 \ MN$

The force of the concrete on the safe = $\underline{29.\overline 3 \ MN}$

Second part:

The gravitational force of the Earth on the safe, W = The weight of the safe

W = Mass, m × Acceleration due to gravity, g

W = 2,200 kg × 9.81 m/s² ≈ 21,582 N

The ratio of the force exerted by the concrete to the weight of the safe is found as follows;

$\displaystyle Ratio \ of \ forces = \frac{29.\overline 3 \times 10^6 \, N}{21,582 \, N} = \frac{4,000,000}{2,943} \approx \mathbf{1359.16}$

The force exerted by the concrete is approximately 1,359.16 times the weight of the safe.

Learn more here:

brainly.com/question/21060171

5 0

3 years ago