All categories

3 years ago

The first step in the process for facturing the trinonmial x2-5x-36 is to

1 answer:

7 0

Set up your parentheses, and put x in the front of both because x is squared. It should look like this: (x )(x ). Next, you find notice that one sign must be positive, and one must be negative. Then, you find the factors of 36. 6 and 6, 3 and 12, 1 and 36, 18 and 2, and 9 and 4. You then guess and check with the different signs and factors what will get you back to the trinomial using the FOIL method. When you do this, you should get (x-9)(x+4).

You might be interested in

What are the four functions that are increasing on the interval

mina [271]

Answer:

Tbh i don’t really know

Step-by-step explanation:

Ask someone else

6 0

2 years ago

Read 2 more answers

What is a rational number between 9.6 and 9.7

lisov135 [29]

I'm pretty sure there are no rational numbers between 9.6 and 9.7

5 0

3 years ago

Read 2 more answers

X and Y are independent random variables. X is normally distributed with mean 1 and variance 1. Y is normally distributed with m

g100num [7]

Answer:

Step-by-step explanation:

Equation of variance:

Total Variance = sum (i=1, N) {c^2,variance^2}

= c1^2*variance1^2 + c2^2*variance2^2 + .....

Variance of X = 1

Variance of Y = 5

Variance of Y - 2X = (1^2)*(5) - (2^2)**(1)

= 5 - 4 = 1

8 0

3 years ago

Tell whether the table of values represents a linear function, an exponential function, or a quadratic funtion

gizmo_the_mogwai [7]

Answer:

Step-by-step explanation:

if it is linear then it will be a straight line(gradient is the same)

if quadratic then curve(gradeint isnt the same)

y=mx+c

m=[y(2)-y(1)]/[x(2)-x(1)]

you can choose any 2 points from the table

m=2-0.4/0-1

m=-1.6

repeat but 2 different coordinates

m=0.4-0.08/-1--2==>-2.24

m=-2.24

different coordinate therefore quadratic

cant be exponential, because nothing is being raised to some power

5 0

2 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Mkey [24]

For this case, we have to:

By definition, we know:

The domain of $f (x) = \sqrt [3] {x}$ is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.

So, we have:

$y = \sqrt [3] {x-2}$ with $x = 0$ : $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x+2}$ with $x = 0:\ y = \sqrt [3] {2}$ is also defined.

$f (x) = \sqrt {x}$ has a domain from 0 to ∞.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .

While $y = \sqrt {x+2}$ if it is defined for $x = 0$ .

Answer:

$y = \sqrt {x-2}$

Option b

6 0

3 years ago