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3 years ago

PLEASE HELP

Mathematics

1 answer:

Maurinko [17]3 years ago

6 0

Hey there! I'm happy to help!

Here is our equation.

$2y-3x=10$

Let's add 3x to both sides.

$2y=3x+10$

Divide both sides by 2.

$y=\frac{3}{2}x+5$

Here is slope intercept form.

$y=mx+b\\m=slope\\b=y-intercept$

So, we can just find those two things in the equation, and here are our answers.

$y=\frac{3}{2}x+5\\m=\frac{3}{2}\\b=5$

The graph is down below. If our y-intercept is 5, then one of our points is (0,5). You can then plug a random x-value into the formula to find another point and then draw the line going through the two points.

$y=\frac{3}{2}(2)+5\\y=3+5\\y=8\\(2,8)$

Have a wonderful day and keep on learning! :D

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We are given polynomial function f(x)= $x^4-x^3-19x^2-x-20$

We are given two zeros -4 and 5.

Therefore, x=-4 and x=5 is given.

So, the two factors of the polynomial will be (x+4)(x-5).

Let us write some steps to factor the given polynomial.

Let us take above factor (x+4) first.

On factoring (x+4) we get factors

$x^4-x^3-19x^2-x-20=\left(x+4\right)\left(x^3-5x^2+x-5\right)$

$\mathrm{:Let \ us \ factor}\:x^3-5x^2+x-5 \ now.$

Grouping

$=\left(x^3-5x^2\right)+\left(x-5\right)$

Factoring out gcf from each group, we get

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So, the final factored form of given polynomial will be

$x^4-x^3-19x^2-x-20=\left(x+4\right)\left(x-5\right)\left(x^2+1\right)$

For first to factors (x+4) and (x-5) we have given roots: -4 and 5.

Let us find the root of third factor we got.

$x^2+1=0$

Subtracting both sides by 1.

x^2 = - 1.

On taking square root on both sides, we get a square root(-1)

$x=\sqrt{-1}=$ + i and -i.

Those are imaginary solutions.

Therefore, correct option is B) No, there are two imaginary solutions.

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3 years ago

￼PLEASE HELP

PLEASE HELP