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lara [203]
2 years ago
7

For the first part of the obstacle course, Helena needs to sprint as fast as she can to Checkpoint A, which is 5 units north of

her starting point.
Which mapping statement correctly describes how she gets from the starting point (0,0) to Checkpoint A?

(x,y)↦(x,y+5)
(x,y)↦(x+5,y)
(x,y)↦(x−5,y+5)
(x,y)↦(x+5,y+5)
Mathematics
1 answer:
andrew11 [14]2 years ago
5 0

Answer:

If the four number sentences at the bottom are the choices for your answer the correct one would be choice A or 1.  

I hope this is what you meant if not I'll try to take my answer down and try to answer it again correctly!

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Answer 2,3,4 for 25 points.
viktelen [127]

Answer:

2 is -1/3

3 is -6

4 is 2

Step-by-step explanation:

2 is 7x + 8x = 3x - 4

3 is 5x + (-3) = 2x + (-15)

4 is -4x + 17 = 6x -3

3 0
3 years ago
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What is C in -43=-5c+4(2c+7)​
elixir [45]

Answer: C = - 71/3 or -23.6 or -23 2/3    All the same

Step-by-step explanation:

4 0
3 years ago
When the area in square units of an expanding circle is increasing twice as fast as its radius in linear units
alexira [117]

Answer:

r=1/π

Step-by-step explanation:

Area of the circle is defined as:

Area = πr²

Derivating both sides

\frac{dA}{dr}=2πr

\frac{dA}{dt}  =  \frac{dA}{dr} x \frac{dr}{dt}  =  2πr\frac{dr}{dt}

If area of an expanding circle is increasing twice as fast as its radius in linear units. then we have : \frac{dA}{dt}  =2\frac{dr}{dt}

Therefore,

2πr \frac{dr}{dt}  =  2  \frac{dr}{dt}

r=1/π

5 0
3 years ago
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Gina bought a thumbdrive and 12 writting pads for $13y . Each writting pad cost$2 Find the cost of the thumbdrive​
elixir [45]

Answer:

its 24 since each one cost 2$

5 0
3 years ago
6. Find the values for each of the following: a. 3 times 5 in Z1o b. 3 times 5 in Zs c. 3 times 5 in Z3 e. 5 3 in Z1o
Anettt [7]

Answer:  The calculations are done below.

Step-by-step explanation:  We are given to find the values for each of the following :

(A) 3 times 5 in \mathbb{Z}_{10}.

We know that

\mathbb{Z}_{10}=\{0,1,2,3,4,5,6,7,8,9\}(\textup{mod}10).

Therefore, we get

3\times5=15(\textup{mod}10)=5.

(B) 3 times 5 in \mathbb{Z}_{5}.

We know that

\mathbb{Z}_{5}=\{0,1,2,3,4\}(\textup{mod}5).

Therefore, we get

3\times5=15(\textup{mod}5)=0.

(C) 3 times 5 in \mathbb{Z}_{3}.

We know that

\mathbb{Z}_{3}=\{0,1,2\}(\textup{mod}3).

Therefore, we get

3\times5=15(\textup{mod}3)=0.

(D) 5 times 3 in \mathbb{Z}_{10}.

We know that

\mathbb{Z}_{10}=\{0,1,2,3,4,5,6,7,8,9\}(\textup{mod}10).

Therefore, we get

5\times3=15(\textup{mod}10)=5.

Thus, the required values are evaluated.

8 0
3 years ago
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