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Natali [406]
3 years ago
13

A dog and a cat are 200 meters apart when they see each other. The dog can run at a speed of 30 m/sec, while the cat can run at

a speed of 24 m/sec.
Chapter Reference

How soon will they meet if they simultaneously start running towards each other?

Mathematics
2 answers:
Deffense [45]3 years ago
7 0
Use the pencil workings

Verizon [17]3 years ago
7 0

Answer:

They will meet in 3.(703) seconds

Step-by-step explanation:

30+24=54

200/54=3.(703)

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Answer:

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7 0
3 years ago
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8 0
3 years ago
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7 8 9 10 11 12 13 14 15 plz that is all but I don't get it and my math teacher sucks so when you tell me the answers can you sho
ohaa [14]
The formula for area is pi x r^2 (pi times radius squared) so, for number 7 youre given the radius or "r", which is half the length of the inside of the cirlce. On a calculator: you would do 8 times 8 (8^2) because that's the radius squared and then you should get 64 and times 64 by 3.14 or use the pi button depending on what your teacher prefers you to use. 

When you're given diameter or "d" which represents the is equal to the radius times 2. So, for example, number 8 the diameter is 26 so, the radius is 13. Use the formula for area just like in number 7 to find the area of this circle. 

Make sure you pay attention to what information youre given (whether its "d", Diameter, "r" or radius. Use those two lists of steps for solving numbers 7 through 12. 

For numbers 13-15:
use the number (22/7) instead of 3.14 or the pi button on your calculator and do the same steps!!!!!

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7 0
3 years ago
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vovangra [49]
I’d say D as the answer
7 0
2 years ago
Use the exponential decay​ model, Upper A equals Upper A 0 e Superscript kt​, to solve the following. The​ half-life of a certai
Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

A=A_0 e^{kt}

Where,

A_0 = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = \frac{A_0}{2},

i.e.

\frac{A_0}{2}=A_0 e^{19k}

\frac{1}{2} = e^{19k}

0.5 = e^{19k}

Taking ln both sides,

\ln(0.5) = \ln(e^{19k})

\ln(0.5) = 19k

\implies k = \frac{\ln(0.5)}{19}\approx -0.03648

Now, if the substance to decay to 78​% of its original​ amount,

Then A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0

0.78 A_0=A_0 e^{-0.03648t}

0.78 = e^{-0.03648t}

Again taking ln both sides,

\ln(0.78) = -0.03648t

-0.24846=-0.03648t

\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0
3 years ago
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