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4 years ago

A B C D What is the correct?

Mathematics

2 answers:

Sindrei [870]4 years ago

8 0

Answer:c i think hopefully its right

Step-by-step explanation:

Hitman42 [59]4 years ago

4 0

Answer:

Step-by-step explanation:

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Divide using a common factor of 6 to find an equivalent fraction for

Savatey [412]

5: six twelfths is one half

6: eight tenths is four fifths

7: four eighths is one half

8: ten tenths is two halves

9: two sixths is one third

10: one quarter, because you divide the top and bottom number by 3

method:

take the first one - four eighths, which is 4/8

the question gives you what to divide them by, 4, so u divide both the top number and bottom number to get 1/2, which is one half :)

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3 years ago

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karen has 60 dollars in a savings account. the interest rate is 10% and is not compounded. how much will she have in 1 year

lisov135 [29]

$66. 10% of $60 is $6, so add that to $60.

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3 years ago

WILL MARK BRAINLYIST IF CORRECT!

Nady [450]

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3 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]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

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

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

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4 years ago

Write an equation that shows the equivalency between meters and gigameters.

Anastasy [175]

Below is an equation that shows the equivalency between meters andGigameters:

$10 ^{9} m = 1 Gm$

The conversion factor should be used to convert from meters toGigameters should be:

$\frac{1 Gm}{10 ^{9} m }$

4 0

4 years ago