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KatRina [158]
3 years ago
12

A line passes through (1, - 1) and (3, 5). What is the equation of the line in slope-intercept form?​

Mathematics
1 answer:
lisabon 2012 [21]3 years ago
8 0

Answer: a+b+c=0 and 3a+5b+c=0

Step-by-step explanation:

If you find a solution to these equations, you will find an equation of your line. Please notice that (a,b,c) defines the same line as (ka,kb,kc) and for any non-zero k, and a=b=c=0 defines no line equation at all. Now you can take a=1 and solve for b and c, or take b=1 and solve for a and c, or take c=1 and solve for a and b, at least one of these will work.

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Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess
Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2

4 0
3 years ago
Read 2 more answers
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dezoksy [38]
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6 0
3 years ago
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Ronch [10]

Answer:

The pairs of functions that best represent the equation are f(x)=x² and g(x)=1/x

Step-by-step explanation:

For this problem, you will have to multiply each equation in the answer choices until you find the pair that has a product of x.

Let's start with the first answer choice.

(f * g)(x)

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Now, we multiply these terms together.

x² * 1/x = x²/x

Now, divide x from x². When you are dividing exponents, the exponential numbers subtract. So, since there is no exponent on x, then we assume this exponent to be 1. So, when dividing you will subtract 1 from 2.

x²/x = x

So, the answer to the problem is answer choice A. These functions multiply together to get you a final product of x.

7 0
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Answer:

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Step-by-step explanation:

that's the answer I thinkkkk

4 0
3 years ago
Pls simplify this for me ​
strojnjashka [21]

Answer:

I think the answer is c but I didn’t simplify bc I don’t know what you mean

Step-by-step explanation:

4 0
3 years ago
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