(2y-9x) ^2 please, I am in a crunch

Mathematics

2 answers:

ikadub [295]3 years ago

4 0

Answer:

www.symbolab.com , its a easy algebra calculator. wish u luck

Step-by-step explanation:

Misha Larkins [42]3 years ago

4 0

(2y-9x)²
(2y-9)(2y-9)
the equation is ² so break it
into two bracket

4y²-18y -18y+81 multiply it out
=4y² - 36y + 81 (the answer)

I assume this would be the correct answer but you could simplify further

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What is bigger 5/6 or 7/12

Fynjy0 [20]

5/6 is the correct answer!

6 0

4 years ago

Which is the following is the set of real zeros of the function f(x) = (x3 + 1000)(x4 - 160,000)?

marusya05 [52]

Answer:

A. { -20, -10, 20 }

Step-by-step explanation:

Given:

The function is given as:

$f(x)=(x^3+1000)(x^4-160000)$

Let us simplify the function.

First, we use the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$

$x^2+1000= x^3+10^3=(x+10)(x^2-10x+10^2)\\x^3+1000=(x+10)(x^2-10x+100)$

Next, we use the identity $a^4-b^4=(a-b)(a+b)(a^2+b^2)$

$x^4-160000=x^4-20^4=(x-20)(x+20)(x^2+20^2)=(x-20)(x+20)(x^2+400)$

Now, the function can be rewritten as:

$f(x)=(x+10)(x^2-10x+100)(x-20)(x+20)(x^2+400)$

Now, the zeros are those values of $x$ for which $f(x)=0$

Now, for $f(x)=0$ , we must have either of the factors 0.

$x+10=0\\x=-10$

$x-20=0\\x=20\\\\x+20=0\\x=-20$

The factors $x^2-10x+100$ and $x^2+400$ can have no zeros as the first one has imaginary roots and second one is always greater than 0 irrespective of the $x$ values.