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Nat2105 [25]
1 year ago
11

On a number​ line, suppose point E has a coordinate of -4 ​, and EG = 13. What are the possible coordinates of point​ G?

Mathematics
1 answer:
Ludmilka [50]1 year ago
7 0

Answer:

9 or -17

Step-by-step explanation:

Hello!

The distance can be positive or negative, as we are just counting the number of units they are apart.

To find the possible coordinates for G, we can add or subtract 13 from -4 to find the possible coordinates for G.

<h3>Find G</h3>
  • -4 + 13 = G1
    -4 - 13 = G2
  • 9 = G1
    -17 = G2

The possible coordinates of G are 9 or -17.

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Answer:

$16

Step-by-step explanation:

48÷3= $16 per pair of shoe

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3 years ago
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Does anyone know the step by step for this 9(m +5) - 3(m - 2) = 8m + 31
tigry1 [53]
<h3>Answer :</h3>

<u>Given</u> :

  • 9(m + 5) - 3(m - 2) = 8m + 31

We have to find value of m.

➠ 9(m + 5) - 3(m - 2) = 8m + 31

➠ 9m + 45 - 3m + 6 = 8m + 31

➠ 9m - 3m + 45 + 6 = 8m + 31

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➠<u> m = 10</u>

<h3>Hope It Helps!</h3>

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3 years ago
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Bus A and Bus B leave the bus depot at 3 pm.
katrin [286]

Step-by-step explanation:

Bus A leave depot at = 3pm

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both leave depot at 3 pm so

Bus A complete its route in25 mins

Bus A reach at his point at 3:25 pm

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2 years ago
In quadratic drag problem, the deceleration is proportional to the square of velocity
Mars2501 [29]
Part A

Given that a= \frac{dv}{dt} =-kv^2

Then, 

\int dv= -kv^2\int dt \\  \\ \Rightarrow v(t)=-kv^2t+c

For v(0)=v_0, then

v(0)=-kv^2(0)+c=v_0 \\  \\ \Rightarrow c=v_0

Thus, v(t)=-kv(t)^2t+v_0

For v(t)= \frac{1}{2} v_0, we have

\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\  \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\  \\ \Rightarrow kv_0t=2 \\  \\ \Rightarrow t= \frac{2}{kv_0}


Part B

Recall that from part A, 

v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\  \\ \Rightarrow dx=-kv^2tdt+v_0dt \\  \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\  \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a

Now, at initial position, t = 0 and v=v_0, thus we have

x=a

and when the velocity drops to half its value, v= \frac{1}{2} v_0 and t= \frac{2}{kv_0}

Thus,

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Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

- \frac{1}{2k} + \frac{2}{k} +a-a \\  \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}
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3 years ago
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-BARSIC- [3]

Answer:

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Step-by-step explanation:

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