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OleMash [197]
2 years ago
13

How many halves are in three wholes

Mathematics
1 answer:
vlabodo [156]2 years ago
5 0

Answer:

6

Step-by-step explanation:

1/2 + 1/2 = 1

1/2 + 1/2 = 1

1/2 + 1/2 = 1

There are 6 halves that equal 3.

Another way to do this is:

1/2x = 3

3 / 1/2 = 6

x = 6

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Solve the following system of equations using Substitution: Submit your answer as an ordered pair. 2x−3y=32x-3y=32x−3y=3 x=y+2x=
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Answer:

  (3, 1)

Step-by-step explanation:

We assume you want the solution to the system ...

  • 2x−3y=3
  • x=y+2

The second equation gives a nice expression for x, so we can use that in the first equation.

  2(y+2) -3y = 3 . . . . substitute for x in the first equation

  2y +4 -3y = 3 . . . . . eliminate parentheses

  -y = -1 . . . . . . . . . . . collect terms, subtract 4

  y = 1 . . . . . . . . . . . . multiply by -1

  x = 1 +2 = 3 . . . . . . substitute for y in the second equation

The solution is (x, y) = (3, 1).

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3 years ago
Express 25 minutes as a fraction of 2 hours in its simplest form
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A smaller triangle is cut out of a larger triangle. The larger triangle is a right triangle with side lengths of 5 millimeters a
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Step-by-step explanation:

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There is a triangle with a perimeter of 63 cm, one side of which is 21 cm. Also, one of the medians is perpendicular to one of t
NemiM [27]

Answer:

21cm; 28cm; 14cm

Step-by-step explanation:

There is no info in the problem/s  text which one of the triangle's  side is 21 cm. That is why we have to try all possible variants.

Let the triangle is ABC . Let the AK is the angle A bisector and BM is median.

Let O is AK and BM cross point.

Have a look to triangle ABM.  AO is the bisector and AOB=AOM=90 degrees (means that AO is as bisector as altitude)

=> triangle ABM is isosceles => AB=AM  (1)

1. Let AC=21   So AM=21/2=10.5 cm

So AB=10.5 cm as well.  So BC= P-AB-AC=63-21-10.5=31.5 cm

Such triangle doesn' t exist ( is impossible), because the triangle's inequality doesn't fulfill.  AB+AC>BC ( We have 21+10.5=31.5 => AB+AC=BC)

2. Let AB=21 So AM=21 and AC=42 .So  BC= P-AB-AC=63-21-42=0 cm- such triangle doesn't exist.

3. Finally let BC=21 cm. So AB+AC= 63-21=42 cm

We know (1) that AB=AM so AC=2*AB.  So AB+AC=AB+2*AB=3*AB

=>3*AB=42=> AB=14 cm => AC=2*14=28 cm.

Let check if this triangle exists ( if the triangle's inequality fulfills).

BC+AB>AC    21+14>28 - correct=> the triangle with the sides' length 21cm,14 cm, 28cm exists.

This variant is the only possible solution of the given problem.

6 0
3 years ago
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