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wel
2 years ago
10

Help me please

Mathematics
1 answer:
aleksley [76]2 years ago
7 0
270 Problems in 3 Hours
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During the mayoral election,two debates were held between the canidates. The first debate lasted 1 4/5 hours. The second one las
Andrews [41]

Answer:

3\frac {6}{25} hrs \ or \ 3 hrs\  14 mins \ 24 sec

Step-by-step explanation:

The question calls requires one to get the product of the given time. Since first debate lasted for :

1\frac {4}{5} \ hrs

-and the second lasted

1\frac {4}{5} hrs times more than the first then the second took then the first step will involve converting the mixed fractions into improper fraction which will be:

\frac {9}{5}

-Now multiplying

\frac {9}{5}\times\frac{9}{5}\\\\=\frac{81}{25}=3\frac{6}{25}hrs

5 0
3 years ago
The area of a parallelogram is 125 square inches and the height of the parallelogram is 10 inches. What is the lenght of the bas
kipiarov [429]

Answer:

12.5 inches is the correct answer.

Step-by-step explanation:

I can bet you with brainliest post if it is correct mark me brainliest and wrong then spam me(;

6 0
3 years ago
It is your birthday and you are going to get a new cell phone. The day you buy the phone, you will have to pay the $49 interest
Juli2301 [7.4K]

Answer:

30.46

Step-by-step explanation:

4 0
3 years ago
4 Tan A/1-Tan^4=Tan2A + Sin2A​
Eva8 [605]

tan(2<em>A</em>) + sin(2<em>A</em>) = sin(2<em>A</em>)/cos(2<em>A</em>) + sin(2<em>A</em>)

• rewrite tan = sin/cos

… = 1/cos(2<em>A</em>) (sin(2<em>A</em>) + sin(2<em>A</em>) cos(2<em>A</em>))

• expand the functions of 2<em>A</em> using the double angle identities

… = 2/(2 cos²(<em>A</em>) - 1) (sin(<em>A</em>) cos(<em>A</em>) + sin(<em>A</em>) cos(<em>A</em>) (cos²(<em>A</em>) - sin²(<em>A</em>)))

• factor out sin(<em>A</em>) cos(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (1 + cos²(<em>A</em>) - sin²(<em>A</em>))

• simplify the last factor using the Pythagorean identity, 1 - sin²(<em>A</em>) = cos²(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (2 cos²(<em>A</em>))

• rearrange terms in the product

… = 2 sin(<em>A</em>) cos(<em>A</em>) (2 cos²(<em>A</em>))/(2 cos²(<em>A</em>) - 1)

• combine the factors of 2 in the numerator to get 4, and divide through the rightmost product by cos²(<em>A</em>)

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - 1/cos²(<em>A</em>))

• rewrite cos = 1/sec, i.e. sec = 1/cos

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - sec²(<em>A</em>))

• divide through again by cos²(<em>A</em>)

… = (4 sin(<em>A</em>)/cos(<em>A</em>)) / (2/cos²(<em>A</em>) - sec²(<em>A</em>)/cos²(<em>A</em>))

• rewrite sin/cos = tan and 1/cos = sec

… = 4 tan(<em>A</em>) / (2 sec²(<em>A</em>) - sec⁴(<em>A</em>))

• factor out sec²(<em>A</em>) in the denominator

… = 4 tan(<em>A</em>) / (sec²(<em>A</em>) (2 - sec²(<em>A</em>)))

• rewrite using the Pythagorean identity, sec²(<em>A</em>) = 1 + tan²(<em>A</em>)

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (2 - (1 + tan²(<em>A</em>))))

• simplify

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (1 - tan²(<em>A</em>)))

• condense the denominator as the difference of squares

… = 4 tan(<em>A</em>) / (1 - tan⁴(<em>A</em>))

(Note that some of these steps are optional or can be done simultaneously)

7 0
2 years ago
Which angle measure is greater?
Sliva [168]

Answer:

A

Step-by-step explanation:

plz tell me if correct and is so plz give brainliest

4 0
2 years ago
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