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Anit [1.1K]
3 years ago
14

Write an equation that is perpendicular to y= 1/2x+5 and passes through (4,8)

Mathematics
1 answer:
goblinko [34]3 years ago
4 0

Answer:

y = -2x + 16.

Step-by-step explanation:

The slope of the perpendicular line = -1 / slope of the given line

= -1 / 1/2 = -2.

Using the point slope form of the equation of a straight line:

y - y1 = m (x - x1)

y - 8 = -2(x - 4)

y - 8 = -2x + 8

y = -2x + 16  is the required equation.

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(x + 12)2 + (y+13)2 = 36 <br><br> What’s the steps to get the center and the radius?
Anon25 [30]

Step-by-step explanation:

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A spinner has 5 equally sized sections, 1 of which is gray and 4 of which are blue. The spinner is spun twice. What is the proba
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Which of the following equations could be th equation to represent the given graph? Make sure you explain your answer thoroughly
Alex17521 [72]

Answer:

Option (C) is correct.

Step-by-step explanation:

The given options of the possible equation for the graph are as follows:

(A) y=2\left(\frac{3}{2}\right)^x \\\\(B) y=-2\left(\frac{3}{2}\right)^{-x} \\\\(C) y=2\left(\frac{2}{3}\right)^x \\\\(D) y=-2\left(\frac{2}{3}\right)^{-x} \\\\

The given graph is decreasing and at x=0, y=2.

So, first checking the value of the given options for x=0

(A) y=2\left(\frac{3}{2}\right)^0=2\times 1= 2 \\\\(B) y=--2\left(\frac{3}{2}\right)^{-0}= -2\times 1= -2 \; (not\; possible) \\\\(C) y=2\left(\frac{2}{3}\right)^0= 2\times 1= 2 \\\\(D) y=2\left(\frac{2}{3}\right)^{-0} = -2\times 1= -2 \; (not\; possible)

As, for x=0, y=2, so options (C) and (D) are not possible, so rejected.

Now, checking the nature (increasing or decreasing) of the given equation by differentiating it.

For option (A),

\frac{dy}{dx}=2\left(\frac{3}{2}\right)^{x}\times \ln\left(\frac{3}{2}\right)

As \ln \left(\frac{3}{2}\right)=\ln(1.5)>0 \;and\; \left(\frac{3}{2}\right)^{x} >0

So, \frac{dy}{dx}>0

Therefore, the function in option (A) is increasing function.

Similarly, for option (C),

\frac{dy}{dx}=2\left(\frac{2}{3}\right)^{x}\times \ln\left(\frac{2}{3}\right)

As \ln \left(\frac{2}{3}\right)=\ln(0.67)0

So, \frac{dy}{dx}

Therefore, the function in option (C) is decreasing function.

As the given graph is decreasing, so, (C)  representsy=2\left(\frac{2}{3}\right)^x the given graph.

Hence, option (C) is correct.

6 0
2 years ago
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