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3 years ago

Maddie has 180 coins. Of these coins, 1/12 are dimes. About how many dimes does she have?

1 answer:

8 0

Remember that "of" means to multiply.
(1/12) of 180 coins are dimes. The number of dimes will be
(1/12)*180 = 180/12 = 15

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Solve the square root equation.

Neko [114]

Answer:

x = 9

Step-by-step explanation:

subtract 5 from 11 with = 6

divide by -2 which = -3

square your answer so (-3)² = 9

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2 years ago

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Nuetrik [128]

8159 divided by five times two then, divide that by twenty and multiply by 13, The answer is about 2611 kids wore helmets properly while riding bike

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3 years ago

Find sin θ, cot θ, and csc θ, where θ is the angle shown in the figure.

Natalka [10]

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2 years ago

Joe's gas tank is 1/4 full. After he buys 6 gallons of gas, it is 5/8 full. How many gallons can Joe's tank hold?

Korolek [52]

Answer:

Amount of gas that can be hold by Joe's tank = 16 gallons.

Step-by-step explanation:

Given that Joe's gas tank is 1/4 full. After he buys 6 gallons of gas, it is 5/8 full. Now we need to find about how many gallons can Joe's tank hold.

So let's find change that happened after filling the gas.

Change = (5/8 full) - (1/4 full)

Change = (5/8 - 1/4) full

Change = (5/8 - 2/8) full

Change = (5 - 2)/8 full

Change = 3/8 full

So that means 3/8 full happens in 6 gallons

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So the amount of gas that can be hold by Joe's tank = 16 gallons.

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3 years ago

Assume you have such a watch. if a minimum of 18.0% of the original tritium is needed to read the dial in dark places, for how m

laila [671]

Given that the half life of 3H is 12.3 years.

The amount of substance left of a radioactive substance with half life of $t_{ \frac{1}{2} }$ after t years is given by

$N(t)=N_0\left(\frac{1}{2} \right)^{ \frac{t}{t_{\frac{1}{2}} }$

Therefore, the number of years it will take for 18% of the original tritinum to remain is given by

$18 \% = 100 \% \left(\frac{1}{2} \right)^{ \frac{t}{12.3}} \\ \\ \Rightarrow\left(\frac{1}{2} \right)^{ \frac{t}{12.3}} =0.18 \\ \\ \Rightarrow\frac{t}{12.3}\ln\left(\frac{1}{2} \right)=\ln0.18 \\ \\ \Rightarrow\frac{t}{12.3}= \frac{\ln0.18}{\ln\left(\frac{1}{2} \right)} = \frac{-1.715}{-0.6931} =2.474\\ \\ \Rightarrow t=2.474(12.3)=30.4$

Therefore, the number of <span>years that the time could be read at night is 30.4 years.</span>

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3 years ago