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valkas [14]
2 years ago
11

What is , Pd? 24 4 1 16

Mathematics
1 answer:
lutik1710 [3]2 years ago
5 0
The answer is 24 the first choice
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2x - 3 = - ( 4x + 9 )
mafiozo [28]

2x-3 = -(4x+9)

2x-3 = -4x-9

2x+4x=3-9

6x=-6

x=-1

3 0
3 years ago
Mike had 119 dollars to spend on 6 books. after buying them he had 11 dollars. how much did each book cost ?
deff fn [24]
18 $ Hope I helped
This is how I came to the result..
119-11= 108
108 : 6 = 18 $
8 0
3 years ago
Theresa Mitsoto invested $3,500 in a 3-year CD that paid 6.5% annual interest. She cashed out the CD at the end of two years wit
Andre45 [30]

Answer:

$56.875

Step-by-step explanation:

Given that :

Amount invested = principal, p = 3500

Interest rate, r = 6.5% = 0.065

Penalty on withdrawal = 3 month simple interest

Simple interest = principal * rate * time

Time = 3 months = 3/12 = 0.25 years

Hence,

Simple interest = 3500 * 0.065 * 0.25

Simple interest = $56.875

Hence, penalty paid = $56.875

7 0
3 years ago
Find the domain and the range of the relation shown in the given graph. Also, determine whether the relation is a function.
Ivahew [28]

Answer:

yes, it is a function

D: {-4, -3, 0, 4}

R: {2, 4}

Step-by-step explanation:

7 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
3 years ago
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