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3 years ago

PLEASE HELP, THE TUTOR OPTION ISNT WORKING

Mathematics

1 answer:

PSYCHO15rus [73]3 years ago

5 0

Answer:

Cost of one rocket = $1.5

Step-by-step explanation:

Given:

6 rockets and 11 bombs cost = $36.50

14 rockets and 8 bombs cost = $41

Find:

Cost of one rocket

Computation:

Cost of one rocket = x

Cost of one bomb = y

So,

6x + 11y = 36.50.....eq1

14x + 8y = 41......eq2

From eq1

y = [36.50 - 6x]/11

From eq2

14x + 8y = 41

14x + 8[36.50 - 6x]/11 = 41

x = 1.5

Cost of one rocket = $1.5

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3 years ago

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!!!!!!!!15 POINTS FOR ANSWERING THIS CORRECTLY!!!!!!!!

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13) If AOBC is a rectangle whose three vertices are A(0,3), (0,0) and B(5,0).

Levart [38]

Answer:

(b) 68

Step-by-step explanation:

Distance between two points:

Suppose that we have two points, $(x_1,y_1)$ and $(x_2,y_2)$ . The distance between them is given by:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Diagonal:

Diagonals are a line between opposite points in a square.

In this question, it is between A(0,3) and B(5,0), and between (0,0) and (5,3). Both these diagonals have the same measure, and are the distance between these points. So

$D = \sqrt{(5-0)^2+(3-0)^2} = \sqrt{5^2 + 3^2} = \sqrt{34}$

Then sum of the squares of its both diagonals is units

Both diagonals measure $\sqrt{34}$ . Sum of squares is:

$(\sqrt{34})^2 + (\sqrt{34})^2 = 34 + 34 = 68$

The correct answer is given by option b.

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3 years ago

(18 - square root of 49 + 4)^2

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Answer:

15^2 = 225

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4 years ago

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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

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3 years ago