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3 years ago

If you have a line that goes through (-1, 7) and (8, -2), what would the equation of the line be in point-slope form?

Mathematics

1 answer:

Eddi Din [679]3 years ago

5 0

Answer:

y-9 = -1/12(x-8)

Step-by-step explanation:

To write an equation of a line perpendicular to the graph of y = 12x-3 and passing through the point, we will follow the following steps.

The standard form of point-slope form of the equation of a line is given as

y − y1 = m(x − x1),

m is the slope of the unknown line

(x1, y1) is a point on the line.

Step 1: We need to calculate the slope of the known line first,

Given y = 12x-3

from the equation, m = 12 on comparison.

Step 2: get the slope of the unknown line. since the line given is perpendicular to the line y = 12x - 3, the product of their slope will be -1 i.e Mm = -1

M = -1/m

M is the slope of the unknown line

M = -1/12

Step 3: We will substitute M = -1/12 and the point (8, 9) into the point-slope form of the equation of a line i.e y − y1 = M(x − x1),

M = -1/12, x1 = 8 and y1 = 9

y-9 = -1/12(x-8)

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prisoha [69]

Answer:

$\boxed{\color{blue}\textsf{The percentage increase in price is \textbf{512.22\% }}}$

Step-by-step explanation:

Here the earlier price of the video game was $ 49 and now it's being sold for $ 300 . We need to calculate the percentage increase in price .

So the increase in cost is $( 300 - 49 ) = $ 251 .

So we can calculate the % increase as ,

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Substituting the respective values :-

$\sf\implies \% Increase = \dfrac{ Increase\ in \ price }{Original\ price } \times 100 \\\\\sf\implies \% Increase =\dfrac{\$ 251}{\$ 49 }\times 100 \\\\\sf\implies \% Increase = 5.122 \times 100 \\\\\implies \boxed{\pink{\frak{ \% Increase = 512.24 \% }}}$

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3 years ago

For what value of c is the function defined below continuous on (-\infty,\infty)?

kozerog [31]

$f(x)= \left \{ {{x^2-c^2,x \ \textless \ 4} \atop {cx+20},x \geq 4} \right
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It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
A function, f, is continuous at x = 4 if
 $\lim_{x \rightarrow 4} \ f(x) = f(4)$

In notation we write respectively
 $\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)$

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just 4c + 20.

On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
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Thus these two limits, the one from above and below are equal if and only if
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That is to say, if c = -2, f(x) is continuous at x = 4.

Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers $(-\infty, +\infty)$

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