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scoray [572]
3 years ago
8

What is the angle of rotation of this shape? Round to the nearest tenth of a degree.

Mathematics
2 answers:
natali 33 [55]3 years ago
4 0
It is 24.0 so A i think 
Lady bird [3.3K]3 years ago
3 0
Its not c so DO NOT PICK C

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Parallel graph of 2x+y=5 y-intercept is (0,4)
-Dominant- [34]

Isolate y

2x + y = 5

y = -2x + 5

The slope of the new line should be -2, since parallel lines have same slope.

Since y-int. is 4:

y = -2x + 4 is the new equation.

5 0
3 years ago
Would it make sense to state "a dilation has a scale factor of 1" explain your reasoning
adelina 88 [10]
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4 years ago
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IrinaK [193]
Question 1:
If you know that 3 squares equals 7 circles and the squares are being counted by 3, then continue to count the circle by 3 until you get 25 circles.
Example:
3 Squares : 7 circles
4 squares : 10 circles
5 squares : 13 circles 
6 squares : 16 circles
7 squares : 19 circles
8 squares : 22 circles
9 squares : 25 circles       Add 9 squares to 3 = 12
You will have 12 squares when continues to count all the way to 25 circles.

Question 2:
Just multiply 14 by 28 and you will get your answer. Which is 3. The missing digit's 3. You can then check by doing 392 ÷ 14 to see if you get 28.
6 0
3 years ago
V(t), left parenthesis, t, right parenthesis models the number of visitors in a park as a function of the outside temperature t
Kruka [31]

Answer:

The number of visitors increases at the same rate over both intervals

Step-by-step explanation:

The unit rate at which the number of visitors in the park increases over a given temperature interval is called the average rate of change, or ARCARCA, R, C.

To find the average rate of change of a function over an interval, we need to take the total change in the function value over the interval and divide it by the length of the interval.

Hint #22 / 3

We are asked to compare the rates at which the number of visitors increases over the interval between an outside temperature of 181818 degrees Celsius and 202020 degrees Celsius, and over the interval between an outside temperature of 202020 degrees Celsius and 272727 degrees Celsius. These correspond to the domain intervals [18,20][18,20]open bracket, 18, comma, 20, close bracket and [20,27][20,27]open bracket, 20, comma, 27, close bracket.

Let's calculate the average rate of change of VVV over those intervals:

ARC_{[18,20]}ARC

[18,20]

​

A, R, C, start subscript, open bracket, 18, comma, 20, close bracket, end subscript ARC_{[20,27]}ARC

[20,27]

​

A, R, C, start subscript, open bracket, 20, comma, 27, close bracket, end subscript

\begin{aligned} \dfrac{V(20)-V(18)}{20-18}&=\dfrac{18-10}{2}\\\\&=\dfrac{8}{2}\\\\&=4\end{aligned}\quad

20−18

V(20)−V(18)

​

​

 

=

2

18−10

​

=

2

8

​

=4

​

 \begin{aligned} \dfrac{V(27)-V(20)}{27-20}&=\dfrac{46-18}{7}\\\\&=\dfrac{28}{7}\\\\&=4\end{aligned}

27−20

V(27)−V(20)

​

​

 

=

7

46−18

​

=

7

28

​

=4

​

Hint #33 / 3

The average rate of change over the interval [18,20][18,20]open bracket, 18, comma, 20, close bracket is the same as the average rate of change over the interval [20,27][20,27]open bracket, 20, comma, 27, close bracket.

Therefore, the number of visitors increases at the same rate over both intervals.

7 0
3 years ago
Solve the following equations: <br> 4 1/5 ÷ 1 1/5 = 11 - 1 7/18x
Savatey [412]

Answer: what kind of equations is that, that is mad hard

7 0
3 years ago
Read 2 more answers
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