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nadya68 [22]
3 years ago
14

On the planet Joop, there are 6 gespils to every 14 vreels. If farmer

Mathematics
1 answer:
zzz [600]3 years ago
3 0

Answer:

51 vreels

Step-by-step explanation:

Step 1: Given data

Ratio of gespils to vreels on the planet Joop: 6:14

Step 2: Calculate the number of gespils per 120 vreels

Elentine has 120 vreels on her rew farm. If there are 6 gespils every 14 vreels, the number of gespils is:

120 vreels × (6 gespils/14 vreels) ≈ 51 vreels

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What is it's momentum?
Aliun [14]

\huge \boxed{ \sf  Solution}

  • Mass of the truck (m) = 30000 Kg
  • Velocity (v) = 25 m/s
  • Let the momentum be p.
  • We know, momentum = mass × velocity
  • Therefore, the momentum of the truck (p)
  • = mv
  • = (30000 × 25) Kg m/s
  • = 750000 Kg m/s

<u>Answers</u><u>:</u>

<u>Variable </u><u>:</u><u> </u><u>m,</u><u> </u><u>v,</u><u> </u><u>p</u>

<u>Equation </u><u>:</u><u> </u><u>p </u><u>=</u><u> </u><u>mv</u>

<u>Answer:</u><u> </u><u>The momentum of the truck is 750000 Kg m/s.</u>

Hope you could understand.

If you have any query, feel free to ask.

5 0
2 years ago
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lesya692 [45]
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5 0
3 years ago
exactly 2/5 of the unversity students live off campus. Of those students, 3/10 of them live at home. what fraction of the unives
jekas [21]
To find the answer, we multiply these two fractions
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7 0
3 years ago
On a typical long distance call you talk for 30 minutes. On a typical local call you talk for 10 minutes. Your phone company off
klio [65]

The number of long distance calls to be made can be obtained by linear programming

  • To minimize the phone bill, the number of long distance call to be made is none

Reason:

Maximum number of number of long distance calls = 15

Maximum number of local calls = 30

Number of minutes of calls to make each month = 240 minutes

Duration of each long distance call = 30 minutes

Duration of each local call = 10 minutes

Cost of long distance call = $0.08

Cost of local call = $0.03

Solution:

Let <em>X</em>, represent the number of long distance calls, and let <em>Y</em> represent the number local calls

The objective function is given as follows;

  • P = 0.08·(30)·X + 0.03·(10)·Y

P = 2.4·X + 0.3·Y

The constraints are;

X ≥ 0

Y ≥ 0

X ≤ 15

Y ≤ 30

30·X + 10·Y ≥ 240

Solving we have;

  • Y ≥ 24 - 3·X

The above inequality can be plotted with MS Excel

From the objective function, P = 2.4·X + 0.3·Y, the coefficient of the long

distance calls, <em>X</em> is larger than the coefficient of the local calls <em>Y</em>, therefore,

making the minimum possible number of long distance calls of 0, and 24

local calls will give a cost;

  • P = 2.4 × 0 + 0.3 × 24 = 7.2

Therefore, to minimize the phone bill, the number of long distance calls to

be made is <u>zero long distance calls</u>

Learn more about linear optimization here:

brainly.com/question/21491539

5 0
3 years ago
A rectangle is 6yards longer than it is wide. Find the dimensions of the rectangle if it’s area is 187 square yards
emmasim [6.3K]

Answer:

Width = 11 yards

Length = 17 yards

Step-by-step explanation:

First of all, the length of the rectangle is 6 yards longer than the width, this means, length = width + 6 yards. This dimensions can be represented on figure 1, where <em>w</em> is width, and <em>l</em>, for length.

We know the area of a rectangle is A = width x length

For our case 187 = w . (w + 6)

Using the Distributive Property for the multiplication we obtain

187 = w^{2} +6w

w^{2} +6w-187 =0,

Using the quadratic formula w=\frac{-b\±\sqrt{b^{2}-4ac } }{2a} where a = 1, b = 6, c = - 187 and replacing into the formula, we will have:

w=\frac{-6\±\sqrt{6^2-4(1)(-187)} }{2(1)}

w=\frac{-6\±\sqrt{36+748} }{2}=\frac{-6\±\sqrt{784} }{2}=\frac{-6\±28}{2}

We have two options: w=\frac{-6+28}{2}=\frac{22}{2}=11  yards

Or

w=\frac{-6-28}{2}=\frac{-34}{2}=-17 yards But a distance (width) can not be negative so, this answer for w must be discarded.

The answer must be width = 11 yards.

To find the length l =\frac{187}{11}=17 yards

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3 years ago
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