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Which number is a solution of the inequality?

Oksana_A [137]

The first one is 6 and the second I is 0

3 0

4 years ago

The length of a photograph of Mr. Lemley playing golf is 1475 inches. If the area

Mice21 [21]

Answer:

1.12 × 10^-3

Step-by-step explanation:

The computation of the width of the photograph is as follows:

As we know that

Area of the rectangle = length × width

33 ÷ 20 square inches = 1475 inches × width

So, the width is

= Length ÷ area

= 1475 inches ÷ (33 ÷ 20 square inches)

= 1.12 × 10^-3

8 0

3 years ago

If 1/3 of 33 bakers can make 89 pretzels in 2 1/2 hours, then how many pretzels can 5 3/4 bakers make in 1 1/2 hours? Explain yo

kirza4 [7]

Answer:

Step-by-step explanation:

11 bakers makes 89 pretzels in 2.5 hours

11 bakers makes 35.6 pretzels in 1 hour

1 baker makes 3.23636 pretzels in 1 hour

so, 5.75 bakers in 1.5 hours at 3/23636 pretzels per hour will make....

5.75* 1.5 * 3.23636 = 27.9136 pretzels

4 0

3 years ago

What is the interquartile range of this data set 10,12,16,

Mademuasel [1]

The answer is 6.

Ignore this I’m just typing for more letters

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