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3 years ago

Functions.

Mathematics

1 answer:

never [62]3 years ago

4 0

Answer:

2,5 and 6

Step-by-step explanation:

The answer is going to be option 2, 5 and 6 because the line touches -4,2 and 4 on the x axis. And these are the x values so that's why this is going to be the answer.

Hope this information helps you!

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Find an equation of the line that (a) has y-intercept of -8 and (b) is parallel to the line 6 x − 1 y = 7 6x-1y=7 . Write your a

faust18 [17]

Answer:

The equation of the line in the slope-intercept form is:

$\:y=6x-8$

Step-by-step explanation:

The slope-intercept form of the equation of a line is

$y\:=\:mx\:+\:b$

As y intercept is -8, we have

$y\:=\:mx\:-8$

Since the line is parallel to $6x-y=7$ , then it has the same slope.

Let's put this equation in $y\:=\:mx\:+\:b$ form and find the slope

$6x-y=7$

$y=6x-7$

Thus, the slope is :

$m = 6$

Now we can finish the equation of the line in the slope-intercept form

$y\:=\:mx\:+\:b$

$\:y=6x-8$ ∵ $m = 6$ , b = -8

Therefore, the equation of the line in the slope-intercept form is:

$\:y=6x-8$

7 0

3 years ago

Pls help me ill give brailiest do very soon

Nikitich [7]

A.6x+34
B.8t+38
C.2y-14
D.11n+16

5 0

3 years ago

Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.

vesna_86 [32]

Factorize the denominator:

$\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}$

If $x\neq\pm2$ , we can cancel the factors of $x^2-4$ , which makes $x=-2$ and $x=2$ removable discontinuities that appear as holes in the plot of $g(x)$ .

We're then left with

$\dfrac1{x+1}$

which is undefined when $x=-1$ , so this is the site of a vertical asymptote.

As $x$ gets arbitrarily large in magnitude, we find

$\displaystyle\lim_{x\to-\infty}g(x)=\lim_{x\to+\infty}g(x)=0$

since the degree of the denominator (3) is greater than the degree of the numerator (2). So $y=0$ is a horizontal asymptote.

Intercepts occur where $g(x)=0$ ( $x$ -intercepts) and the value of $g(x)$ when $x=0$ ( $y$ -intercept). There are no $x$ -intercepts because $\dfrac1{x+1}$ is never 0. On the other hand,

$g(0)=\dfrac{0-4}{0+0-0-4}=1$

so there is one $y$ -intercept at (0, 1).

The domain of $g(x)$ is the set of values that $x$ can take on for which $g(x)$ exists. We've already shown that $x$ can't be -2, 2, or -1, so the domain is the set

$\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}$

8 0

3 years ago

Can someone help me with this question please.

galina1969 [7]

Answer:

6, 8, and 10.

Step-by-step explanation:

You could work this out with the pythagorean theorem, by proving that 6^2, 36, plus 8^2, 64, equals 100. The fastest way, however, is to use pythagorean triples. These are predetermined sets of numbers that work as side lengths for right triangles. The first two are 3, 4, and 5, which form a right triangle, and 6, 8, and 10, shown here.

5 0

2 years ago

The table gives the scores of 6 students from a class of 25 in a competitive exam. the point estimate of the mean score for the

Ymorist [56]

The mean of the scores if 6 students in a competitive exam is 25 marks.

Given scores of 6 students are: 10,30,50,40,20 in a competitive exam.

We have to find out the mean of the mark of 6 students in a competitive exam.

Mean is the sum of all the values in a set of data, such as numbers, measurements, divided by the number of values. It is also known as average.

Mean=∑X/n

∑X=10+30+50+40+20

=150

N=6

Now to calculate the mean we have to divide 150 by number of students which is 6.

Mean=150/6

=25 marks.

Hence the mean of the scores of 6 students is 25 marks.

Learn more about mean at brainly.com/question/1136789

#SPJ4

Question is incomplete as it should include :

Marks of students =10,30,50,40,20.

8 0

1 year ago