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Dafna1 [17]
3 years ago
6

Question nr9 needs to be solved

Mathematics
1 answer:
Ksivusya [100]3 years ago
8 0

No9.

let Abby =2x, Brenda=x and Carl=1/4x

2x+x+1/4x=48.75

3x+1/4x =48.75

3x+x/4 =48.75

LCM=4

We have, 13x/4 =48.74

cross multiply

13x = 195

x= 15

hence, Brenda receive $15

Abby receives 2×15=$30

and Carl received 15/4=$3.75

check: 15+30+3.75= $48.75

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What’s the equivalent expression for -2 (-9x + 6y)
Strike441 [17]

Answer:

the expression would be: 18x-12y

Step-by-step explanation:

8 0
3 years ago
The numerical value of sin²5° + sin²10° + sin²15° +... sin²85° + sin²90° is equal a) 17/2 b) 19/2 c) 15/2 d) 13/2​
Vikki [24]

Answer: b)~\Large\boxed{\frac{19}{2} }

Step-by-step explanation:

<h3>Given expression</h3>

sin²5° + sin²10° + sin²15° +... sin²85° + sin²90°

<h3>Concept:</h3>

sin²x + cos²x = 1

sin(x) = cos (90 - x)

<u />

<u>There are in total these terms:</u>

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90

<u>In total, there are 18 terms, and the first one matches with the second to the last one:</u>

5 -- 85

10 -- 80

.

.

.

40 -- 50

<u />

<u>There are 2 terms left over:</u>

sin²45 and sin²90

<h3>Convert the first half of the sine terms (sin²5 - sin²40) to the cosine terms</h3>

sin²5 = cos² (90 - 5) = cos²85

sin²10 = cos² (90 - 10) = cos²80

.

.

.

sin²40 = cos² (90 - 40) = cos²50

<h3>Simplify the 16 grouped terms </h3>

<em>i.e. sin²85 and cos²85</em>

Using the concept of sin²x + cos²x = 1

sin²85 + cos²85 = 1

sin²80 + cos²80 = 1

.

.

.

sin²50 + cos²50 = 1

Total = (16/2) × 1 = 8 × 1 = 8

<h3>Evaluate the 2 terms that are left over</h3>

sin²45 = (sin45) (sin45) = (√2 / 2) (√2 / 2) = 1/2

sin²90 = (sin90) (sin 90) = (1) (1) = 1

<h3>Add all the terms together</h3>

8+\dfrac{1}{2} +1=\Large\boxed{\frac{19}{2} }

Hope this helps!! :)

Please let me know if you have any questions

8 0
1 year ago
Please help me i posted this 4 times already no one answered
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