Find two numbers that multiply to make the 1st number and add to make the 2nd number. Multiply -42; Add 1 _, and _

Mathematics

1 answer:

dedylja [7]2 years ago

5 0

Answer:

-6 and 7

Step-by-step explanation:

It is useful to look at the factor pairs of -42.

-42 = -1(42) = -2(21) = -3(14) = -6(7)

Sums of these factor pairs are 41, 19, 11, 1.

You want the last pair of factors: -6 and 7.

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Which expression is equivalent to square root 55x^7y6/11x^11y^8

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Answer:

$\large\boxed{\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\dfrac{\sqrt5}{x^2y}}$

Step-by-step explanation:

$\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\sqrt{\dfrac{55}{11}\cdot\dfrac{x^7}{x^{11}}\cdot\dfrac{y^6}{y^8}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=\sqrt{5x^{7-11}y^{6-8}}=\sqrt{5x^{-4}y^{-2}}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt5\cdot\sqrt{x^{-4}}\cdot\sqrt{y^{-2}}=\sqrt5\cdot\sqrt{x^{(-2)(2)}}\cdot\sqrt{y^{(-1)(2)}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt5\cdot\sqrt{(x^{-2})^2}\cdot\sqrt{(y^{-1})^2}\qquad\text{use}\ \sqrt{a^2}=a$

$=\sqrt5\cdot x^{-2}\cdot y^{-1}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\to a^{-1}=\dfrac{1}{a}\\\\=\sqrt5\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}=\dfrac{\sqrt5}{x^2y}$