What is 3k(-5k+21)+2(2.5k+90) <-- Confused please help

1 answer:

5 0

Use distributive prpoerty
a(b+c)=ab+ac

distribute each
3k(-5k+21)=-15k^2+63k

2(2.5k+90)=5k+180

so nowe we have
-15k^2+63k+5k+180
add like terms
-15k^2+68k+180
that is the answer

You might be interested in

If f(x) = 2x^3 - 14x^2 + 38x -26 and x-1 is a factor of f(x) find all of the zeros of f(x) algebraically.

Anestetic [448]

Answer:

Step-by-step explanation:

First confirm that x = 1 is one of the zeros.

f(1) = 2(1)^3 - 14(1)^2 + 38(1) - 26

f(1) = 2 - 14 + 38 - 26

f(1) = -12 + 38 = + 26

f(1) = 26 - 26

f(1) = 0

=========================

next perform a long division

x -1 || 2x^3 - 14x^2 + 38x - 26 || 2x^2 - 12x + 26

2x^3 - 2x^2

===========

-12x^2 + 28x

-12x^2 +12x

==========

26x -26

26x - 26

========

Now you can factor 2x^2 - 12x + 26

2(x^2 - 6x + 13)

The discriminate of the quadratic is negative. (36 - 4*1*13) = - 16

So you are going to get a complex result.

x = -(-6) +/- sqrt(-16)

=============

x = 3 +/- 2i

f(x) = 2*(x - 1)*(x - 3 + 2i)*(x - 3 - 2i)

The zeros are

3 +/- 2i

8 0

3 years ago

What is the quotient 2y^-6y-20/4y+12 ÷ y+5y+6/3y^2+28y+27

malfutka [58]

Answer with explanation:

$\rightarrow \frac{\frac{2y^2-6 y-20}{4 y+12}}{\frac{y^2+5 y+6}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{y^2-3y-10}{2 y+6}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{(y-5)(y+2)}{2 (y+3)}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{(y-5)(y+2)}{2 (y+3)}} \times {\frac{3 y^2+28 y+27}{(y+2)(y+3)}}\\\\ \rightarrow\frac{(y-5)\times(3 y^2+28 y+27)}{2 (y+3)^2}}$