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exis [7]
3 years ago
10

For the graph, locate the x-intercept and the y-

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
7 0
X intercept= 6
y intercept= -2
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Which mathematical operation is the inverse of addition
Amiraneli [1.4K]
Additive inverse is the number multiplied with -1 such that the sum of the number and the additive inverse will result in 0.
7 0
3 years ago
. What is the negative solution to this equation?<br> 8x2 – 4x = 363
stiv31 [10]

Answer:

- x = 81 and 3/4

Step-by-step explanation:

8 × 2 – 4x = 363

16 - 4x = 363

Subtract 16 from each side:

16 - 16 - 4x = 363 - 16

- 4x = 327

Divide each side by 4:

- 4x  ÷ 4 = 327 ÷ 4

- x = 81 and 3/4

8 0
3 years ago
Define linear relationship​
-Dominant- [34]

Answer:

A linear relationship (or linear association) is a statistical term used to describe a straight-line relationship between two variables. Linear relationships can be expressed either in a graphical format or as a mathematical equation of the form y = mx + b. Linear relationships are fairly common in daily life.

Step-by-step explanation:

5 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

4 0
3 years ago
Read 2 more answers
This is distance &amp; midpoint formulas I don’t know how to solve this can anyone help?
Tasya [4]

Answer:

midpoint formula: (x₁ + x₂)/2, (y₁ + y₂)/2

distance: √[(x₂ - x₁)² + (y₂ - y₁)²]

Step-by-step explanation:

What points are you trying to calculate the distance and the midpoint for?

8 0
3 years ago
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