What is 4 3/7 written as an improper fraction?

Mathematics

2 answers:

aleksley [76]3 years ago

5 0

31/7
First you multiply the denominator(7) by the whole number(4) then you would add the numerator(3) to get the new numerator place the new numerator(31) I’ve the old denominator(7)
I hope this helps

Katyanochek1 [597]3 years ago

4 0

Hello,

4 3/7 would be written as 19/7

Because in order to get that you would need to do 4x3 plus 7 and use the same denominator.

I hope this helps if it did please make me brainliest!

I attached a picture if you still need help! :D

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Troyanec [42]

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3 0

3 years ago

Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab

Aliun [14]

Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

$\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$