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3 years ago

−3 1/3 ÷ 9= What I need it now!

Mathematics

1 answer:

Contact [7]3 years ago

8 0

Step 1. Convert 3 1/3 to improper fraction.

- 3 * 3 + 1/3 ÷ 9

Step 2. Simplify 3 * 3 to 9

-9 + 1/3 ÷ 9

Step 3. Simplify 9 + 1 to 10

-10/3 ÷ 9

Step 4. Use this rule: a ÷ b/c = a * c/b

-10/3 * 1/9

Step 5. Use this rule: a/b * c/d = ac/bd

-10 * 1/3 * 9

Step 6. Simplify 10 * 1 to 10

-10/3 * 9

Step 7. Simplify 3 * 9 to 27

-10/27

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<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B2x%7D%20%2B%2012%20%3D%20x%20%2B%208%20" id="TexFormula1" title=" \sqrt{2x} + 12

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First note that $x\ge0$ ; otherwise $\sqrt{2x}$ is undefined as long as we're only looking for real solutions.

We can write

$\sqrt{2x}+12=x+8$
$\sqrt{2x}=x-4$
$2x=(x-4)^2$
$2x=x^2-8x+16$
$x^2-10x+16=0$
$(x-8)(x-2)=0$

$\implies x=8,x=2$

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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

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Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

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

<u>Right part:</u>

$\dfrac{1}{\sin^2A-\sin^4A}\\ \\=\dfrac{1}{\sin^2A(1-\sin^2A)}\\ \\=\dfrac{1}{\sin^2A\cos^2A}$

Since simplified left and right parts are the same, then the equality is true.

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