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3 years ago

A stone dropped from a bridge strikes the water 2.2 s later. How high is the bridge above thewater

1 answer:

7 0

Since you are giving no information whatsoever, the formula could be:

S = (1/2)g.t², where s is the distance, 9 =9.8 m/s² and t = time
S = (1/2).(9.8)(2.2)² ans S= 23.7 meters

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Solve the problem.

yarga [219]

Answer:

39%

Step-by-step explanation:

Let's find the answer by using the following formula:

selling price=(paid price)+((paid price)*(percent markup)) which can be written as:

percent markup=((selling price)-(paid price))/(paid price)

percent markup=(27.80-20)/20

percent markup=0.39=39%

In conclusion, the percent markup is 39%.

4 0

3 years ago

Kay's mother taught her how to make handmade ornaments to sell at a craft fair. Kay rented a table at

omeli [17]

Answer: B

Step-by-step explanation: 6x represents the positive because it represents the amount being made from each ornament. The -30 represents the one-time cost of the table. And the -2.50x represents the cost of each ornament.

3 0

3 years ago

The half-life of a certain tranquilizer in the bloodstream is 22 hours. How long will it take for the drug to decay to 90% of

nasty-shy [4]

Answer:

It will take approximately 3.34 hours for the drug to decay to 90% of the original dosage

Step-by-step explanation:

As suggested, we use the formula for exponential decay:

$A(t)=A_0\,e^{-k\,t}$

From the given information, the half life of the drug in blood id 22 hours, so that means that it takes that number of hours to go from the initial value $A_0$ , to a final value equal to $A_0/2$ . Using this information we can find the decay rate "k" by solving for this parameter in the formula, and using the natural log function to bring the exponent down:

$A(t)=A_0\,e^{-k\,t}\\\frac{A_0}{2} =A_0\,e^{-k\,*22}\\\frac{A_0}{A_0*2} =e^{-k\,*22}\\\frac{1}{2} =e^{-k\,*22}\\ln(\frac{1}{2})=-k\,*22\\ k=-\frac{ln(\frac{1}{2})}{22} \\k=0.0315$

Now we use this value for the decay rate "k" to calculate how long it would take to decay to 90% of the original dose;

$A(t)=A_0\,e^{-0.0315\,t}\\0.9*A_0} =A_0\,e^{-0.0315\,t}\\\frac{0.9*A_0}{A_0} =e^{-0.0315\,t}\\0.9 =e^{-0.0315\,t}\\ln(0.9)=-0.0315\,t\\ t=-\frac{ln(0.9)}{0.0315} \\t=3.3447\,hours$

7 0

3 years ago

Louis and Max are contestants in a jellybean-eating contest. Louis eats 18 jellybeans

Nezavi [6.7K]

Answer: They ate at the same pace.

Step-by-step explanation: assuming you want to know who ate more faster you can do 18 / 30 and 24 / 4 to find out who ate more in 10 seconds each.

18 / 3 = 6 and 24 / 4 = 6. So they ate at the same speed / pace.

Hope this helps! Brainliest and / or a rating out of 5 stars would be appreciated :)

8 0

3 years ago

If a truck averages 12 miles to the gallon, how many pounds of carbon dioxide are emitted into the atmosphere when the truck tra

77julia77 [94]

Answer: 278.30 pounds (if the gasoline does not contain ethanol)

317.12 pounds (if the gasoline does contain ethanol)

Step-by-step explanation:

If the truck averages 12 miles per gallon, then in 170 miles it consumes:

170mi/12mi = 14.17 gallons.

We know that for one gallon consumed, the carbon dioxide emitted into the atmosphere is 19.64 pounds. (assuming it does not contain ethanol)

Then for 14.17 gallons consumed, we have a emission of:

14.17*19.64 pounds = 278.30 pounds

I the gasoline contains ethanlol, for a gallon the emiision is around 22.38 pounds.

In this case the total emission is:

14.17*22.38 pounds = 317.12 pounds

3 0

3 years ago