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alekssr [168]
3 years ago
13

Find the midpoint between two points on a number line if one of the points is at -7, and the other point is at 12.

Mathematics
1 answer:
frosja888 [35]3 years ago
3 0

Answer:

D

Step-by-step explanation:

The midpoint is the average of the 2 endpoints, that is

midpoint = \frac{-7+12}{2} = \frac{5}{2} = 2.5 → D

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Find the area of a regular nonagon with a side length of 7 and an apothem of 5
vitfil [10]
A NONA=9, nonagon has 9 sides, so it's perimeter is 7+7+7+7+7+7+7+7+7, or 63, and we know the apothem.

bearing in mind that the area of a regular polygon is 
(1/2)ap   a = apothem, p = perimeter

so in this case that'd be (1/2)(5)(63).
7 0
3 years ago
Which type of function is shown in the table below?​
lora16 [44]
Answer: Exponential table
3 0
3 years ago
What is the slope of the line passing through(-7, 8) (-8, 8) is?
Hunter-Best [27]

Answer:

Step-by-step explanation:

(-7, 8) (-8, 8)

(y2 - y1) / (x2 - x1)

(8 - 8) / (-8 + 7)

0 / -1

the slope is 0

you could also see this because the y value doesn't change

if the y value doesn't change the slope is 0, if the x value doesn't change the slope is undefined

3 0
2 years ago
Read 2 more answers
Solve the 3 × 3 system shown below. Enter the values of x, y, and z. X + 2y – z = –3 (1) 2x – y + z = 5 (2) x – y + z = 4 (3)
Svet_ta [14]
<h2>Answer:</h2>

\boxed{x=1, \y=-1, \ z=2}

<h2>Step-by-step explanation:</h2>

We will use the Gaussian elimination method to solve this problem. To do so, let's follow the following steps:

Step 1: Let's multiply first equation by −2. Next, add the result to the second equation. So:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\~~ x&-~~~~~ y&+~~~~ z&~=~4\end{array}

Step 2: Let's multiply first equation by −1. Next, add the result to the third equation. Thus:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&-~~~3~ y&+~~2~ z&~=~7\end{array}

Step 3: Let's multiply second equation by −35, Next, add the result to the third equation. Therefore:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&&+~~\frac{ 1 }{ 5 }~ z&~=~\frac{ 2 }{ 5 }\end{array}

Step 4: solve for z, then for y, then for x:

\frac{ 1 }{ 5 } ~ z & = \frac{ 2 }{ 5 } \\ \\ \boxed{z & = 2}

-5y+3z &= 11\\-5y+3\cdot 2 &= 11\\ \\ \boxed{y &= -1}

By substituting y=-1 \ and \ z=2 into the first equation, we get the x. So:

x+2(-1)-2=-3 \\ \\ x-2-2=-3 \\ \\ \boxed{x=1}

6 0
4 years ago
Read 2 more answers
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