Through: (1, -3), slope = -4<br> A) y=-x+4 B) y = x +4<br> C) y = 4x + 1 D) y=-4x + 1

If you run 10 feet in 3 seconds how long does it take you to run a mile?

mel-nik [20]

It would take you 15 minutes

3 0

2 years ago

Find angles EFG pleaseee

Bumek [7]

Answer:

50 degrees.

Step-by-step explanation:

Angle FGE = $180-(7x+18)$ because angles on a straight line add up to 180 degrees.

Angle EFG = $180-38-(180-(7x+18))=6x-10$ (Angles in a triangle add up to 180)

$180-38-\left(-7x+162\right)=6x-10$

$180-38+7x-162=6x-10$

$7x-20=6x-10$

Add 20 to both sides:

$7x-20+20=6x-10+20$

$7x=6x+10$

Subtract 6x from both sides:

$7x-6x=6x+10-6x$

$x=10$

Angle $EFG=6x-10=6(10)-10=60-10=50$

3 0

3 years ago

Read 2 more answers

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.

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

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

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