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3 years ago

Which is the correct fair share for the

Mathematics

1 answer:

Fantom [35]3 years ago

7 0

It is 1/3
Bc there is 3 c

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A train leaves a station and travels north at a speed of 105105 km/h. twotwo hours later, a second train leaves on a parallel

hram777 [196]

Let x = hours to meet

train one has a 2 hour head start or 210 km head start

210 + 105x = 135x
30x = 210
x = 7 hours to meet

6 0

3 years ago

Stewart is playing a video game. For each prize captured, he earns more points. Which is the dependent variable in the

luda_lava [24]

Don’t mine me just commenting so I can ask a question

5 0

3 years ago

I need with number 16

Arturiano [62]

Answer:

Ben will carry 63/8 or 7.9 of recycling.

Step-by-step explanation:

14 + 10 1/2 + 16 1/4 = 163/4 - 25 = 63/4 ÷ 2 = 63/8 or 7.9

Hope that helps :)

4 0

3 years ago

Which expresses 54 : 81 in its simplest form?

Bingel [31]

The answer would be 2:3.

So basically two - thirds.

Hope this helps and if it does happy studying.

5 0

3 years ago

A substance with a half life is decaying exponentially. If there are initially 12 grams of the substance and after 70 minutes th

77julia77 [94]

Answer: 233 min

Step-by-step explanation:

This problem can be solved by the following equation:

$A=A_{o} e^{-kt}$ (1)

Where:

$A=7 g$ is the quantity left after time $t$

$A_{o}=12 g$ is the initial quantity

$t=70 min$ is the time elapsed

$k$ is the constant of decay for the material

So, firstly we need to find the value of $k$ from (1) in order to move to the next part of the problem:

$\frac{A}{A_{o}}=e^{-kt}$ (2)

Applying natural logarithm on both sides of the equation:

$ln(\frac{A}{A_{o}})=ln(e^{-kt})$ (3)

$ln(\frac{A}{A_{o}})=-kt$ (4)

$k=-\frac{ln(\frac{A}{A_{o}})}{t}$ (5)

$k=-\frac{ln(\frac{7 g}{12 g})}{70 min}$ (6)

$k=0.00769995 min^{-1}$ (7) Now that we have the value of $k$ we can solve the other part of this problem: Find the time $t$ for $A=2 g$ .

In this case we need to isolate $t$ from (1):

$t=-\frac{ln(\frac{A}{A_{o}})}{k}$ (8)

$t=-\frac{ln(\frac{2 g}{12 g})}{0.00769995 min^{-1}}$ (9)

Finally:

$t=232.697 min \approx 233 min$

5 0

3 years ago