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3 years ago

What is the equation of a line passes thru the point (4, 2) and is perpendicular to the line whose equation is y = ×/3 - 1 ??

Mathematics

2 answers:

yawa3891 [41]3 years ago

7 0

Answer:

y = -3x + 14

Step-by-step explanation:

First find the reciprocal slope since it is perpendicular. Slope of the other line is 1/3 so the slope for our new equation is -3.

Plug information into point-slope equation

(y - y1) = m (x-x1)

y - 2 = -3 (x-4)

Simplify if needed

y - 2 = -3x + 12

y = -3x + 14

cupoosta [38]3 years ago

5 0

Answer:

Perpendicular lines have slopes that are opposite and reciprocal. Therefore, the line we are looking for has a -3 slope.

y= -3x+b

Now, we can substitute in the point given to find the intercept.

2= -3(4)+b

2= -12+b

b=14

Finally, put in everything we've found to finish the equation.

y= -3x+14

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The sales tax on a used car is $174, and the sales tax rate is 5%. Find the purchase price (the price before taxes are adde

rosijanka [135]

Answer:

Step-by-step explanation:

Let x be the purchase price

5% of x = $ 174

$\frac{5}{100}*x=174\\\\x=174*\frac{100}{5}\\\\x=174*20\\$

Purchase price = $ 3480

8 0

3 years ago

F:[1/2,infinity)—>[3/4,infinity) <br> f(x)=x^2-x+1 <br> find the inverse of f(x);please explain

dexar [7]

Answer:

$f^{-1}(x)=\frac{1}{2}+\sqrt{x-\frac{3}{4}}$

Step-by-step explanation:

y=x^2-x+1

We want to solve for x.

I'm going to use completing the square.

Subtract 1 on both sides:

y-1=x^2-x

Add (-1/2)^2 on both sides:

y-1+(-1/2)^2=x^2-x+(-1/2)^2

This allows me to write the right hand side as a square.

y-1+1/4=(x-1/2)^2

y-3/4=(x-1/2)^2

Now remember we are solving for x so now we square root both sides:

$\pm \sqrt{y-3/4}=x-1/2$

The problem said the domain was 1/2 to infinity and the range was 3/4 to infinity.

This is only the right side of the parabola because of the domain restriction. We want x-1/2 to be positive.

That is we want:

$\sqrt{y-3/4}=x-1/2$

Add 1/2 on both sides:

$1/2+\sqrt{y-3/4}=x$

The last step is to switch x and y:

$1/2+\sqrt{x-3/4}=y$

$y=1/2+\sqrt{x-3/4}$

$f^{-1}(x)=1/2+\sqrt{x-3/4}$

8 0

3 years ago

Assume that f is continuous on [-4,4] and differentiable on (-4,4). The table gives some values of f'(x) x: -4, -3, -2, -1, 0, 1

kondaur [170]

$f$ will be increasing on the intervals where $f'(x)>0$ and decreasing wherever $f'(x)$ . Local extrema occur when $f'(x)=0$ and the sign of $f'(x)$ changes to either side of that point.

$f'(x)$ is positive when $x$ is between -4 and some number between -2 and -1, and also 2 (exclusive) and 4, so you can estimate that $f(x)$ is increasing on the intervals [-4, -2] and (2, 4].

$f'(x)$ is negative when $x$ is between some number between -2 and -1, up to some number less than 2. So $f(x)$ is decreasing on the interval [-1, 1].

You then have two possible cases for extrema occurring. The sign of $f'(x)$ changes for some $x$ between -2 and -1, and again to either side of $x=2$ .