10y-50= 10y−50= \,\,-20 −20

Mathematics

1 answer:

densk [106]2 years ago

6 0

Answer:

Therefore the value of 'y' is,

$y=3$

Step-by-step explanation:

Given:

$10y-50=-20$

To Find:

y= ?

Solution:

$10y-50=-20$ ........Given

Step 1. Adding 50 to both the side we get

$10y-50+50=-20+50\\10y=30$

Step 2. Dividing by 10 on both the side we get

$\dfrac{10y}{10}=\dfrac{30}{10}\\\\y=3$

Therefore the value of 'y' is,

$y=3$

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The quotient of (x^3+3x^2-4x-12)/(x^2+5x+6)

Elis [28]

Answer: The quotient is (x-2).

Step-by-step explanation:

Since we have given that

$f(x)=(x^3+3x^2-4x-12)\\\\and\\\\g(x)=x^2+5x+6\\\\So,\ \frac{\left(x^3+3x^2-4x-12\right)}{\left(x^2+5x+6\right)}$

Now, we have to find the quotient of the above expression.

So, here we go:

$Factorise\ (x^3+3x^2-4x-12)\\\\=\left(x^3+3x^2\right)+\left(-4x-12\right)\\\\=-4\left(x+3\right)+x^2\left(x+3\right)\\\\=\left(x+3\right)\left(x^2-4\right)$

Now, we will divide the above simplest form with g(x):

$\frac{\left(x+3\right)\left(x^2-4\right)}{\left(x+2\right)\left(x+3\right)}\\\\=\frac{x^2-4}{x+2}\\\\=\frac{\left(x+2\right)\left(x-2\right)}{x+2}\ using\ (a^2-b^2)=(a+b)(a-b)\\\\=x-2$