Find the value of the expression given below. 2/3-1/7

Mathematics

1 answer:

vivado [14]4 years ago

6 0

Answer:

the answer is 11/21 or 0.523809524

hope this helped

Step-by-step explanation:

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Write an equation for the line parallel to the given line that contains C. C (1,8); 5/7x + 7

Genrish500 [490]

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Answer: $\textsf{y = 5/7x + 51/7}$

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Given: $\textsf{Goes through (1, 8) and is parallel to y = 5/7x + 7}$

Find: $\textsf{Write an equation that follows that criteria}$

Solution: We know that our equation is going to parallel to the line that was given therefore the slope would stay the same at 5/7. We also have a point so we can plug in the values into the point-slope form, distribute, and solve for y.

Plug in the values

$\textsf{y - y}_1\textsf{ = m(x - x}_1\textsf{)}$

$\textsf{y - 8 = 5/7(x - 1)}$

Distribute

$\textsf{y - 8 = (5/7 * x) + (5/7 * (-1))}$

$\textsf{y - 8 = 5/7x - 5/7}$

Add 8 to both sides

$\textsf{y - 8 + 8 = 5/7x - 5/7 + 8}$

$\textsf{y = 5/7x - 5/7 + 8}$

$\textsf{y = 5/7x + 51/7}$

Therefore, the final equation that follows the description that was provided in the problem statement is y = 5/7x + 51/7.

4 0

2 years ago

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.

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

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

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