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svetoff [14.1K]
3 years ago
14

A triangle has one side length of 30 and another side length of 10. Select all of the possible lengths of the third side.

Mathematics
1 answer:
marishachu [46]3 years ago
4 0

20,28, and 50 are possible length.

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Explain what their first step would be if they wanted to eliminate the y variable from the equations.
Vinvika [58]

Answer:

  • r1 +4×r3 ⇒ r1
  • r2 -3×r3 ⇒ r2

Step-by-step explanation:

We note that the coefficient of y in equation (iii) is 1, so it is convenient to use that equation to eliminate y from the others.

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<em>To eliminate the y-variable from equation (i), four times equation (iii) can be added to equation (i).</em>

<em>To eliminate the y-variable from equation (ii), three times equation (iii) can be subtracted from equation (i).</em>

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The result of these operations would be ...

  10x +6z = 28\quad\text{(iv)}\\\\-7x-9z=2\quad\text{(v)}

5 0
2 years ago
A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Three h
neonofarm [45]

Answer:

  • 50 ft by 75 ft
  • 3750 square feet

Step-by-step explanation:

Let x represent the length of the side not parallel to the partition. Then the length of the side parallel to the partition is ...

  y = (300 -2x)/3

And the enclosed area is ...

  A = xy = x(300 -2x)/3 = (2/3)(x)(150 -x)

This is the equation of a parabola with x-intercepts at x=0 and x=150. The line of symmetry, hence the vertex, is located halfway between these values, at x=75.

The maximum area is enclosed when the dimensions are ...

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That maximum area is 3750 square feet.

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<em>Comment on the solution</em>

The generic solution to problems of this sort is that half the fence (cost) is used in each of the orthogonal directions. Here, half the fence is 150 ft, so the long side measures 150'/2 = 75', and the short side measures 150'/3 = 50'. This remains true regardless of the number of partitions, and regardless if part or all of one side is missing (e.g. bounded by a barn or river).

6 0
2 years ago
Neptunes average distance from the sun is 4.503x10^9 km. Mercurys average distance from the sun is 5.791x10^7 km. About how many
LuckyWell [14K]
Neptune is 77.76 times farther from the Sun than Mercury is. 


6 0
3 years ago
Solve for x: x - 10 = -12<br><br> A. 2<br> B. -2<br> C. 22<br> D. -22
Sindrei [870]
Add 10 to both sides. X is 2. A.
7 0
3 years ago
Read 2 more answers
An Someone plz help me plz :(
zlopas [31]

Answer:

18

Step-by-step explanation:

original 3:4

now 18:24

Multiply by 6

6 0
2 years ago
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