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pychu [463]
2 years ago
12

Help me on my fractions theres a photo ​

Mathematics
2 answers:
BabaBlast [244]2 years ago
8 0

Answer:

10/23

Step-by-step explanation:

simply ez

nadya68 [22]2 years ago
5 0

Answer:

Step-by-step explanation:

-we need to change the mix number fractions in improper fractions

-find the common denominator so we can add the fractions

-simplify the fractions whenever possible

-convert it back in a mix number

Obs.

Remember that the order of operations requires us to do multiplication first and then addition.

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Answer:

y=4x+35

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2 years ago
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Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3
omeli [17]
\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}

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=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R
3 0
3 years ago
Which proportion could be used to find 77% of 126?.
ss7ja [257]
The answer is D hope it help
5 0
3 years ago
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A square has a diagonal with the length of 8V7 inches. What is the length of the<br> sides?
Nikitich [7]

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3 years ago
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Step-by-step explanation:

Mark me brainliest plzzz

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