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igomit [66]
3 years ago
13

What is the equation of a line that is parallel to -2x+3y=-6 and passes through the point (-2,0)

Mathematics
1 answer:
mr_godi [17]3 years ago
4 0
First, a line that is parallel, means a line that has the same slope as the original. To find the slope of the original equation, we have to solve for y.
-2x+3y=-6
3y=2x-6
y=2/3x-2
From this equation, we can see that the slope of the line is 2/3. For every 2 units you go up, you move three units over.

Now we need to use the point (-2,0) to find the equation of the parallel line. 
y-y=m(x-x)

Plug in the point coordinates and the slope, and solve for the final equation of the line.
y-0=2/3(x+2)
y=2/3x+ 4/3
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A(4,-7) B(-2,1) find AB
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\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{4}~,~\stackrel{y_1}{-7})\qquadB(\stackrel{x_2}{-2}~,~\stackrel{y_2}{1})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[-2-4]^2+[1-(-7)]^2}\implies AB=\sqrt{(-2-4)^2+(1+7)^2}\\\\\\AB=\sqrt{(-6)^2+8^2}\implies AB=\sqrt{36+64}\implies AB=\sqrt{100}\implies AB=10

8 0
3 years ago
Help.........................<br><br>​
garri49 [273]
<h3>Answer: Choice A</h3>

x^2\left(\sqrt[4]{x^2}\right)

=====================================================

Explanation:

The fourth root of x is the same as x^(1/4)

I.e,

\sqrt[4]{x} = x^{1/4}

The same applies to x^10 as well

\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4}

Multiply the exponents 10 and 1/4 to get 10/4

\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4} = x^{10*1/4} = x^{10/4}

\sqrt[4]{x^{10}} = x^{10/4}

-----------------------

If we have an expression in the form x^(m/n), with m > n, then we can simplify it into an equivalent form as shown below

x^{m/n} = x^a\sqrt[n]{x^b}

The 'a' and 'b' are found through dividing m/n

m/n = a remainder b

'a' is the quotient, b is the remainder

-----------------------

The general formula can easily be confusing, so let's replace m and n with the proper numbers. In this case, m = 10 and n = 4

m/n = 10/4 = 2 remainder 2

We have a = 2 and b = 2

So

x^{m/n} = x^a\sqrt[n]{x^b}

turns into

x^{10/4} = x^2\sqrt[4]{x^2}

which means

\sqrt[4]{x^{10}} = {x^2} \sqrt[4]{x^2}

7 0
3 years ago
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