Answer:
Explanation:
<u>1) Calculate the volume of the water in the tank.</u>
- Area of the base of the tank: B = 7 ft × 4 ft = 28 ft²
- Height of the tank: H = 9 in = 9 in × 1 ft / 12 in = 0.75 ft
- Volume of the tank: V = area of the base × height = B × H = 28 ft² × 0.75 ft = 21 ft³.
<u>2) Calculate the weight of 21 ft³ of water.</u>
Since this is not a chemistry question but a math question, I will not use the fomula of density but set a proportion with one unknown:
- 62.4 lb / 1 ft³ = x / 21 ft³
Solve for x:
- x = 21 ft³ × 64 lb / ft³ = 1,310.4 lb.
So, rounding to the next integer, the water in the tank weighs 1,310 pounds, when it is full.
I’m tired I’ve so much homework to complete and a long day of school :(
A^2+b^2=c^2 so c^2=12^2+16^2 which then simplifies to c^2=144+256 then simplify that to c^2=400. After that take the square root of c^2 and the square root of 400. So your answer is c=20
Answer:
First mechanic worked 20 hours and second mechanic worked 5 hours
Step-by-step explanation:
Let the number of hours the first mechanic worked = 
Let the number of hours the second mechanic worked = 
Therefore, we can write 2 equations and then solve them simultaneously:
Rearranging the first equation: 
and substituting into the second equation to find :
Now sub 
into the first equation to find
Therefore, first mechanic (a) worked 20 hours and second mechanic (b) worked 5 hours
At least 4 means its minimum value is 4 so n≥4
