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Gwar [14]
3 years ago
15

Given: y || 2 Prove: m25+ m22 + m26 = 180°

Mathematics
1 answer:
Maurinko [17]3 years ago
7 0

Answer:

1. Statement: y || z

Reason: given

2. Statement: m∠1 + m∠2 + m∠3 = m∠LAM

Reason: angle addition postulate

3. Statement: m∠1 + m∠2 + m∠3 =180

Reason: def. of a straight angle

4. Statement: ∠1 ≅∠5

Reason: alternate interior angles theorem

5. Statement: ∠3 ≅∠6

Reason: alternate interior angles theorem

6. Statement: m∠1 = m∠5

Reason: def. of ≅

7. Statement: m∠3 = m∠6

Reason: def. of ≅

8. Statement: m∠5 + m∠2 + m∠6 = 180

Reason: substitution

Step-by-step explanation:

Hope this helps! :)

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Please help thank you!
Arte-miy333 [17]

Answer:

  • golf: $10
  • batting: $2.50

Step-by-step explanation:

We can write two equations in the two unknowns using the given relations. Let g and b represent the costs of a round of golf and a turn in the batting cage, respectively.

  5g +4b = 60 . . . . . Sylvester's expense

  3g +6b = 45 . . . . . Lin's expense

Dividing the second equation by 3 gives ...

  g +2b = 15   ⇒   2b = 15 -g

Substituting into the first equation, we have ...

  5g +2(2b) = 60

  5g +2(15 -g) = 60 . . . . . substitute for 2b

  3g = 30 . . . . . . . . . subtract 30, collect terms

  g = 10 . . . . . . . divide by 3

__

  2b = 15 -10 = 5 . . . . use the value of g to find b

  b = 2.5 . . . . . . . . divide by 2

Mini golf costs $10 per round; batting cages cost $2.50 per turn.

7 0
2 years ago
The floor of a triangular room has an area of 32 1/2 sq.m. If the triangle’s altitude is 7 172 m, write an equation to determine
sineoko [7]

Answer:

\frac{(2)A}{h} =b

b=0.00906m

Step-by-step explanation:

Hello! To solve this exercise we must remember that the area of ​​any triangle is given by the following equation

A=\frac{bh}{2}

where

A=area=32.5m^2

h=altitude=7172m

b=base

Now what we should do take the equation for the area of ​​a rectangle and leave the base alone, remember that what we do on one side of the equation we must do on the other side to preserve equality

A=\frac{bh}{2} \\\frac{2}{h} A=\frac{bh}{2} \frac{2}{h} \\

\frac{A(2)}{h} =b

solving

\frac{2(32.5)}{7172} =0.0090[tex]\frac{A(2)}{h} =b\\b=0.00906m

4 0
3 years ago
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