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xxMikexx [17]
3 years ago
5

Find the equation of the line that is perpendicular to y = 1/4x – 2 and passes though the point (5, –2). A) y = –1/4x + 18 B) y

= –1/4x – 22 C) y = –4x + 18 D) y = –4x – 22
Mathematics
1 answer:
Mashcka [7]3 years ago
7 0

Since our line is perpendicular to this line, the slope of

our line is the negative reciprocal of 1/4, which is -4/1.

Remember, negative reciprocal is just a fancy way

of saying flip the fraction and change the sign.

So the slope of our line is -4/1 and we use this slope along

with our given point to write the equation of our line.

Start with the point-slope formula.

Y - y1 = m(x - x1).

~Substitute

Y - -2 = -4/1(x - 5).

Minus a negative is plus a positive.

Y + 2 = -4/1x + 20.

Subtract 2 from both sides.

Y = -4/1x + 18

So, our answer is C.

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How much is 5/9 of -3/5?<br><br> A) 1/2<br> B) -1/3<br> C) 1/7<br> D) -25/27
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B. Determine the values of a and b so that f is continuous.<br><br> Please help!
wlad13 [49]

Answer:

following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

\lim_{x\to 2}+f(x) = \lim_{x\to 2}+ (ax^2-bx+3) = 4a-2b+3\\\\ \to  4a-2b+3 = 4\\\\  \therefore \ \ 4a-2b = 1\\\\

The one-sided limits of F(x) at x = 3 must be equivalent for f(x) to be continuous at x = 3.

\to \lim_{x\to 3}- f(x) = \lim_{x\to 3}- (ax^2-bx+3) = 9a-3b+3\\\\ \to \lim_{x\to 3}+ f(x) = \lim_{x\to 3}+ (2x-a+b) = 6-a+b\\\\  

So,

\to 9a-3b+3 = 6-a+b\\\\\therefore\\ \to 10a -4b = 3

\to 4a - 2b = 1......(a)\\\to 10a - 4b = 3.......(b)\\

In equation a multiply the by -2 and then add in the equation b:

\to -8a + 4b = -2\\\to 10a - 4b = 3\\ \to 2a = 1\\\\ \to   a = \frac{1}{2}\\\\ \to \ 4(\frac{1}{2}) - 2b = 1\\\\ \to 2 - 2b = 1\\\\ \to   -2b = -1\\\\ \to b = \frac{1}{2}  

So, the value of a \ and \ b= \frac{1}{2}

4 0
3 years ago
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