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mojhsa [17]
3 years ago
7

Which is the equation of a line that is parallel to the line represented by y= 7/6 x + 4 ?

Mathematics
2 answers:
Nesterboy [21]3 years ago
8 0

In order for two lines to be parallel, they must have the dame slope. In this case, it is 7/6. So the new line Will have the next form:

y = 7/6x + b

Now, assuming that the new line passes through the point (1,1):

1 = 7/6*1 + b

b = -0.17

Replacing b:

y = 7/6x - 0.17

natta225 [31]3 years ago
3 0

Let’s rewrite the given equation

y=7/6x+4

Knowing that the new equation is parallel to this one will only tell us the slope. If it is parallel, it shares the slope.

So far our equation of the new line will be this:

y=7/6x+b

Now we need b. B can be any number other than positive 4. Usually, one would find this through a point. Since you do not have one, we cannot find a definite answer. However, you can substitute any number other than 4 to make this equation parallel.

Hope this helps comment below for more questions :)
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Answer:

Points on the graph:

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Just input the first 4 values for x:

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(2, y)

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3 years ago
Each of 7 students reported the number of movies they saw in the past year. Here is what they reported.
rusak2 [61]

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Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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Andre45 [30]

Answer:

Part A:

c = .86*p

c =1.72 for 2 lbs

Part B:

c =.81p

Step-by-step explanation:

Part A:

Total cost = cost per pound * number of pounds

c = .86*p

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c = .86*2

c =1.72

Part B:

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5 0
3 years ago
The point (0, 0) is a solution to which of these inequalities?
adell [148]

Since - 5 is less than -4, hence the point (0, 0) is a solution to the inequality y-5<3x-4

<h3>Solution to inequality expression</h3>

Inequality are equations not separated by an equal sign. Given the inequality

y-5<3x-4

Check if (0, 0) is a solution

0 - 5 < 3(0) - 4

-5 < -4

Since - 5 is less than -4, hence the point (0, 0) is a solution to the inequality y-5<3x-4

learn more on inequality here: brainly.com/question/24372553

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7 0
2 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\}

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3 years ago
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