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MArishka [77]
3 years ago
5

1 : What postulate or theorem can be used to prove △PQR ~ △STU ??

Mathematics
1 answer:
Alika [10]3 years ago
6 0

Answer:

#1 is the AA postulate

Step-by-step explanation:

It is the AA postulate because both triangles have congruent angles, making the triangles similar.

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Your neighbor needs to put a fence around her front yard and a seperate fence around her backyard. Her front yard is 65 feet lon
elena-14-01-66 [18.8K]
Front yard
65+65+45+45=220
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45+45+35+35=160
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3 years ago
What is m∠2?<br> What is m∠1?
Greeley [361]

Answer:

<2 = 75

<1 = 105

Step-by-step explanation:

<2 and 75 are alternate interior angles and since the lines are parallel, they are equal

<2 = 75

<2 + <1 = 180    since the angles form a line

75+ <1 = 180

Subtract 75 from each side

<1 = 180-75

<1 = 105

6 0
3 years ago
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Determine the value of x in the given diagram, when a║b.
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155 degrees is the answer
7 0
3 years ago
A box of 8 waffles weigh 26.9 ounces.The box weighs 4.5 ounces. What is the weight of each waffle?
Phoenix [80]
Each waffle weighs 5.97 ounces hope this helps!
8 0
3 years ago
Verify:<br>cos(2A)=(cotA-tanA)/cscAsecA​
kolbaska11 [484]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

cot A = \frac{cosA}{sinA}, tanA = \frac{sinA}{cosA}, cscA  = \frac{1}{sinA}, secA = \frac{1}{cosA}

Consider the right side

\frac{cotA-tanA}{cscAsecA}

= \frac{\frac{cosA}{sinA}-\frac{sinA}{cosA}  }{\frac{1}{sinA}.\frac{1}{cosA}  }

= \frac{\frac{cos^2A-sin^2A}{sinAcosA} }{\frac{1}{sinAcosA} }

= \frac{cos^2A-sin^2A}{sinAcosA} × sinAcosA ( cancel sinAcosA )

= cos²A - sin²A

= cos2A

= left side ⇒ verified

5 0
3 years ago
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