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Alexxx [7]
3 years ago
15

Two boxes of apples contain a total of 24 apples. If you halve the number of apples in the first box and add 4 apples to the sec

ond box, the total changes to 20 apples. How many apples are in each box initially?
A. 12 and 12

B. 18 and 6

C. 16 and 8

D. 10 and 14

Mathematics
2 answers:
maks197457 [2]3 years ago
7 0

Step-by-step explanation:

let the no.s of apples in the one box be=x

then no.s of apples in the 2nd box =24-x

according to the question,

24-x+4=20

-x= 20-28

-x=-8

x= 8

no.s of apples in one box= 8

and no.s of apples in another box=24-8= 16

option C

Lena [83]3 years ago
5 0

Answer: option C is the correct answer.

Step-by-step explanation:

Let x represent the number of apples in the first box.

Let y represent the number of apples in the second box.

Two boxes of apples contain a total of 24 apples. It means that

x + y = 24

If you halve the number of apples in the first box and add 4 apples to the second box, the total changes to 20 apples. This means that

x/2 + y + 4 = 20

Cross multiplying by 2, it becomes

x + 2y + 8 = 40

x + 2y = 40 - 8

x + 2y = 32- - - - - - - - - - - - - -1

Substituting x = 24 - y into equation 1, it becomes

24 - y + 2y = 32

- y + 2y = 32 - 24

y = 8

x = 24 - y = 24 - 8

x = 16

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Sinx = 1/2, cosy = sqrt2/2, and angle x and angle y are both in the first quadrant.
Leviafan [203]

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Step-by-step explanation:

we know that

tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}

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sin^{2}(\alpha)+cos^{2}(\alpha)=1

step 1

Find cos(X)

we have

sin(x)=\frac{1}{2}

we know that

sin^{2}(x)+cos^{2}(x)=1

substitute

(\frac{1}{2})^{2}+cos^{2}(x)=1

cos^{2}(x)=1-\frac{1}{4}

cos^{2}(x)=\frac{3}{4}

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step 2

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tan(x)=sin(x)/cos(x)

substitute

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step 3

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we have

cos(y)=\frac{\sqrt{2}}{2}

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sin^{2}(y)+cos^{2}(y)=1

substitute

sin^{2}(y)+(\frac{\sqrt{2}}{2})^{2}=1

sin^{2}(y)=1-\frac{2}{4}

sin^{2}(y)=\frac{2}{4}

sin(y)=\frac{\sqrt{2}}{2}

step 4

Find tan(y)

tan(y)=sin(y)/cos(y)

substitute

tan(y)=1

step 5      

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tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}

substitute

tan(x+y)=[1/\sqrt{3}+1}]/[{1-1/\sqrt{3}}]=3.73

7 0
3 years ago
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