Answer:
You have a problem cheater, Daliah Cobb
Step-by-step explanation:
On a typical number line, we have basically two directions that we can move on. Either to the left where the numbers are negative, of to the right where the numbers are positive.
Given that point E has a coordinate of 1 on a number line, and we are told that the distance between E and another point on the number line is 11 (EG), the possible coordinates of point G are two (either we are moving to the right or to the left). Therefore, possible coordinates of G are:
1 - 11 = -10 (to the left)
1 + 11 = 12 (to the right
Equation of a line:

m = gradient: The difference between two y points and two x points.

c = y-intercept: Where the line crosses the y-axis (x=0)
You have:

so you are missing the m and the c.
To calculate m find two y coordinates -you have (12,
<u>7</u>) and (0, <u>
1</u>)- and subtract them. Then divide this by the subtracted values of the x coordinates -you have (<u>
12</u>, 7) and (<u>
0</u>, 1)- This gives:



To calculate the c, you just see where the line crosses the y-axis. Because you have the point (0, 1), you know that when x=0, y=1. Because x=0 is on the y-axis, you can tell that the line passes through y=1. This makes your c = 1:

When you plug these values into the equation you get your answer:
A geometry app shows lengths that add up to 27.4 units.
The Pythagorean theorem can be used to find the lengths of the segments. Each length is the square root of the sum of squares of the difference in x-coordinates and the difference in y-coordinates.
segment AB: length = √(3²+8²) = √73 ≈ 8.544
segment BC: length = √(7²+4²) = √65 ≈ 8.062
segment CA: length = √(10²+4²) = √116 ≈ 10.770
Then the total perimeter is the sum of these lengths:
... 8.544 +8.062 +10.770 = 27.376 ≈ 27.4 . . . . units
Answer:
Yes
Step-by-step explanation:
To make a triangle the 2 smaller sides must join together to become greater than or equal to the biggest side.
8+7=15
<em>Therefore,</em>
<em>side lengths 8, 7, and 15 do create a triangle.</em>
<em />
<em>Hope this helps </em>