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lbvjy [14]
4 years ago
11

Ax+c=r Solve for x Need help can anybody help?

Mathematics
2 answers:
Sphinxa [80]4 years ago
7 0

Answer:

x = \frac{r-c}{a}

Step-by-step explanation:

To find x you need to isolate x.

First you need to isolate the variable term, or the term with x in it, which in this case is ax. To do this you subtract c from each side of the equation. This will give you the equation: ax = r-c.

Next you have completely isolate x. Because x is being multiplied by a, you have to divide each side of the equation by a in order to isolate x. This give you the equation: x = \frac{r-c}{a} which is the solution.

Hope this helps!

kkurt [141]4 years ago
4 0
What are the numbers for the other ones?
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Which classification describes AMNO with vertices M(2, -3), N(3, 1), and<br> 0(-3, 1)?
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Option D

Step-by-step explanation:

32). Given vertices of the triangle are M(2, -3), N(3, 1) and O(-3. 1).

Distance between two points (x_1,y_1) and (x_2,y_2) is given by the expression,

Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Distance between M(2, -3) and N(3, 1) will be,

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      = \sqrt{17}

Distance between M(2, -3) and O(-3, 1),

MO = \sqrt{(2+3)^2+(-3-1)^2}

      = \sqrt{25+16}

      = \sqrt{41}

Distance between N(3, 1) and O(-3, 1),

NO = \sqrt{(3+3)^2+(1-1)^2}

      = 6

Condition for right triangle,

c² = a² + b² [Here c is the longest side of the triangle]

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MO² = MN² + NO²

(\sqrt{41})^2=(\sqrt{17})^2+6^2

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False.

Therefore, given triangle is not a right triangle.

Since, length of all sides are not equal, given triangle will be a scalene triangle.

Option D is the correct option.

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