All categories

3 years ago

Complete each step in the following solution.

Mathematics

1 answer:

Eduardwww [97]3 years ago

6 0

Please see the attached picture for full solution

hope it helps

Good luck on your assignment

You might be interested in

20 points!! Type the correct answer in the box. Use numerals instead of words. What value of n makes the equation true? -1/5n+7=

vredina [299]

Hey there! :)

Answer:

$\huge\boxed{n = 25}$

-1/5n + 7 = 2

Start by subtracting 7 from both sides:

-1/5n + 7 - (7) = 2 - (7)

-1/5n = -5

Multiply both sides by the reciprocal of -1/5, or -5.

(-5) · (-1/5n) = (-5) · (-5)

n = 25

6 0

3 years ago

10. 9. Solve the equation (x-2)(x+2) = 0 (4F) A. 02,0 B. O-2, 2 C.O-2 D. 2

masha68 [24]

Answer:

x=2 or x=-2

Step-by-step explanation:

I'm not sure what you're asking, but when you solve, set both sets of parentheses equal to 0 and solve. This makes x either equal to 2 or -2.

4 0

3 years ago

The ratio of areas between two similar triangles is 1:4. if one side of the smaller triangle is 2 units, find the measure of the

dimaraw [331]

1:4
smaller side = 2
bigger side = 8 since 2/8 = 1/4
1/4 = 2/x
x=4*2=8

5 0

3 years ago

Find the x- and y-intercepts of the graph of each equation. (set x & y equal to 0; two different

bogdanovich [222]

X= 2 y=4 these are the intercepts

5 0

3 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.

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

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

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

4 0

3 years ago

Read 2 more answers