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anzhelika [568]
3 years ago
5

Hey wats 2x22+264 hejdhdldiedheidjeijdiendidnie

Mathematics
2 answers:
Masja [62]3 years ago
6 0

Answer:

hhahahahahahah

Step-by-step explanation:

9966 [12]3 years ago
4 0

Answer:

The correct answer is 308

Hoped I helped

can u give me a brainliest

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5 5/30 is the answer to the question.
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Which ratio is equivalent to 4:5?
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8:10 and 12: 15

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2 years ago
Plsss help (Cuz i'm dumb:D) Calculate the surface area of the prism below, shape below
marin [14]
Surface area of the the rectangles:
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Surface area of the two triangles:
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5 0
3 years ago
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}

=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1

=\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
Mrs, O'Neil has 2 times as many markers as colored pencils. The total number of markers and colored pencils is 18. How many mark
Natali [406]
36 I believe sorry if it’s wrong what I did was multiply 2x 18 have I nice day good luck!
5 0
3 years ago
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