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2 years ago

Factor out the polynomial 12m^2-m-35

1 answer:

6 0

Hey there!!

Okay here is how you solve this... first you look for factors of 12 and 35... for example...

12m² - m - 35
(4m - 7)(3m + 5) <---- This is the factored form!

m = 7/4 and -5/3


I hope this helps! ^__^

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Find the domain and range for the following functions:

Andreyy89

Answer with Step-by-step explanation:

Since we have given that

a. f: ℝ → ℝ given by f(x) = x + 2

Domain would be all real numbers i.e. (-∞, ∞)

Range would be all real numbers i.e. (-∞, ∞)

b. g: ℝ → ℝ given by g(x) = √x − 1

Domain would be all positive numbers i.e. [0,∞)

Range would be [-1,∞)

c. f(g(x))

$f(g(x))=f(\sqrt{x}+1)=\sqrt{x}+1+2=\sqrt{x}+3$

Domain would be [0,∞)

Range would be [3,∞)

d. g(f(x))

$g(x+2)=\sqrt{x+2}-1$

Domain would be [-2,∞)

Range would be (0,∞)

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3 years ago

Why does the liquid in a thermometer rise on warm days and drop on cold days

abruzzese [7]

Because when the atoms in the liquid are coldthe squeeze altogeter and it makethembgo down and when ther hot the spread out and make them go up

5 0

2 years ago

In this diagram, BAC – EDF. If the

puteri [66]

The triangles are congruent. So.

Triangle ABC is 24 inch^2 and BC is 4 inches.
The area of a triangle is 0.5 x base x height. The will be 12 because 12 x 4 x 0.5 which equals 24.

Triangle DEF is half the size of ABC because bc is 4 and EF is 2 showing it’s half. This then means the height will be half which then mean it will be 6.

This then means we can use the equation to work it out. So. 6 x 0.5 x 2 = 6in^2

The answer is 6inches^2

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2 years ago

The line $y = 3$ intersects the graph of $y = 4x^2 + x - 1$ at the points $A$ and $B$. The distance between $A$ and $B$ can be w

raketka [301]

Answer:

Step-by-step explanation:

Let's find the points $A$ and $B$ .

We know that the $y$ -coordinates of both are $3$ .

So let's first solve:

$3=4x^2+x-1$

Subtract 3 on both sides:

$0=4x^2+x-1-3$

Simplify:

$0=4x^2+x-4$

I'm going to use the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , to solve.

We must first compare to the quadratic equation, $ax^2+bx+c=0$ .

$a=4$

$b=1$

$c=-4$

$\frac{-1 \pm \sqrt{1^2-4(4)(-4)}}{2(4)}$

$\frac{-1 \pm \sqrt{1+64}}{8}$

$\frac{-1 \pm \sqrt{65}}{8}$

Since the distance between the points $A$ and $B$ is horizontal. We know this because they share the same $y-coordinate$ .This means we just need to find the positive difference between the $x$ -values we found for the points of $A$ and $B$ .