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Mathematics

1 answer:

iren2701 [21]3 years ago

3 0

Answer: 2x^2 - 18

BTW: the "•" stands for multiply! and the explanations say what you have to do to get there not the next step.

Steps:
(x-3)(2x+6) Set up for solving
x•2x + 6x - 3•2x - 3•6 Use Foil to multiply
2x^2 + 6x - 6x -18 Calculate the product
2x^2 - 18

You might be interested in

If r and s are positive integers, is \small \frac{r}{s} an integer? (1) Every factor of s is also a factor of r. (2) Every prime

Yuri [45]

Answer:

If statement(1) holds true, it is correct that $\small \frac{r}{s}$ is an integer.

If statement(2) holds true, it is not necessarily correct that $\small \frac{r}{s}$ is an integer.

Step-by-step explanation:

Given two positive integers $r$ and $s$ .

To check whether $\small \frac{r}{s}$ is an integer:

Condition (1):

Every factor of $s$ is also a factor of $r$ .

$r \geq s$

Let us consider an example:

$s = 5^2 \cdot 2\\r = 5^3 \cdot 2^2$

$\dfrac{r}{s} = \dfrac{5^3\cdot2^2}{5^2\cdot2} = 10$

which is an integer.

Actually, in this situation $s$ is a factor of $r$ .

Condition 2:

Every prime factor of s is also a prime factor of r.

(But the powers of prime factors need not be equal as we are not given the conditions related to powers of prime factors.)

Let

$r = 2^2\cdot 5\\s =2^4\cdot 5$

$\dfrac{r}{s} = \dfrac{2^3\cdot5}{2^4\cdot5} = \dfrac{1}{2}$

which is not an integer.

So, the answer is:

If statement(1) holds true, it is correct that $\small \frac{r}{s}$ is an integer.

If statement(2) holds true, it is not necessarily correct that $\small \frac{r}{s}$ is an integer.

8 0

3 years ago

ALGEBRA 2 PLEASE HELP! QUICKLY!! I AM TERRIBLE AT ALGEBRA AND I HAVE LOADS OF STUFF TO Do TODAY CAN YOU PLEASE HELP ME WITH THIS

Genrish500 [490]

A) the posible roots of the function f(x)=x⁴-3x³-5x²+21x+22 are:
=-1,+1,-2,+2,-11,+11,-22+22.

b)
one zero of the funtion can be: x=-1 ⇒y=0 (-1,0)

c)
other zero of the funtion is: x=-2 ⇒y=0 (-2,0)

d)
if x=0 then y=22 (0,22)

e) In the file is the graph.