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miss Akunina [59]
3 years ago
11

Oh my goodness please help its confusing.

Mathematics
1 answer:
Irina18 [472]3 years ago
3 0

Probably the easiest way to do these is to convert them to slope intercept form by solving for y.  When we have y=mx+b, we read off the slope m.

-5x + 2y = 10

Add 5x to both sides,

2y = 5x + 10

Divide both sides by 2,

y = (5/2) x + (10/2)

Obviously 10/2=5 but we don't care about that for this problem.  We read off the slope as

Answer: 5/2, last choice


12 = 4x - 6y

Adding 6y and subtracting 12,

6y = 4x - 12

Dividing by 6,

y = (4/6) x - (12/6)

y = (2/3) x - 2

Answer: 2/3


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6 0
2 years ago
If f(x)=x+7 and g(x)= 1 divided by x-13, what is the domain of (f•g)(x)?
andrew11 [14]
<h2>Hello!</h2>

The answer is:

The domain of the function is all the real numbers except the number 13:

Domain: (-∞,13)∪(13,∞)

<h2>Why?</h2>

This is a composite function problem. To solve it, we need to remember how to composite a function. Composing a function consists of evaluating a function into another function.

Composite function is equal to:

f(g(x))=(f\circ} g)(x)

So, the given functions are:

f(x)=x+7\\\\g(x)=\frac{1}{x-13}

Then, composing the functions, we have:

f(g(x))=\frac{1}{x-13}+7\\

Therefore, we must remember that the domain are all those possible inputs where the function can exists, most of the functions can exists along the real numbers with no rectrictions, however, for this case, there is a restriction that must be applied to the resultant composite function.

If we evaluate "x" equal to 13, the denominator will tend to 0, and create an indetermination since there is no result in the real numbers for a real number divided by 0.

So, the domain of the function is all the real numbers except the number 13:

Domain: (-∞,13)∪(13,∞)

Have a nice day!

5 0
3 years ago
Please help me I need assistance
lawyer [7]

(\stackrel{x_1}{2}~,~\stackrel{y_1}{-15})\qquad (\stackrel{x_2}{14}~,~\stackrel{y_2}{r}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{r}-\stackrel{y1}{(-15)}}}{\underset{run} {\underset{x_2}{14}-\underset{x_1}{2}}}~~ = ~~\stackrel{\stackrel{m}{\downarrow }}{\cfrac{3}{4}}\implies \cfrac{r+15}{12}~~ = ~~\cfrac{3}{4}\implies r+15 = \cfrac{12\cdot 3}{4} \\\\\\ r+15=9\implies \boxed{r=-6}

3 0
2 years ago
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