Answer:
C
Step-by-step explanation:
Area of a square equals to the square of its side, therefor the side of a given area for a square is the square root of the area.

Answer:
The quantity of water in the tank after 15 days is 1610.0 gallons OR 1.61 × 10³ gallons.
Step-by-step explanation:
The amount of water in the tank after 15 days is given by the series
910+(−710)+810+(−610)+⋯+310+(−110)+210
From the series, we can observe that, if water is added for a particular day then water will be drained the following day.
Also, for a day when water is to be added, the quantity to be added will be 100 gallon lesser than the quantity that was last added. Likewise, for a day when water is to be drained, the quantity to be drained will be 100 gallons lesser than the quantity that was last drained.
Hence, we can complete the series thus:
910+(−710)+810+(−610)+710(-510)+610(-410)+510(-310)+410(-210)+310+(−110)+210
To evaluate this, we get
910-710+810-610+710-510+610-410+510-310+410-210+310-110+210
= 1610.0 gallons
Hence, the quantity of water in the tank after 15 days is 1610 gallons OR 1.61 × 10³ gallons.
Yes. 95 is correct.
You have three congruent "indentations" in the right hand side. Thus each section must be 15/3 = 5 cm long.
The top rectangle will be 8*5 = 40 cm^2
The bottom rectangle will also be 8*5 = 40 cm^2
The middle area will be 5(8-5) = 5 * 3 = 15 cm^2
40 + 40 + 15 = 95
Answer:
Ic² + b²l = 13 units.
Step-by-step explanation:
We have to evaluate the expression Ic² + b²l with unknowns b and c and having the values of b and c respectively - 3 and - 2.
Now, Ic² + b²l
= I(- 2)² + (- 3)²l {Putting the values of b and c}
= I4 + 9l
= I13l
= 13 units.
Therefore, Ic² + b²l = 13 units. (Answer)
Answer:
85%
Step-by-step explanation:
To find the percentage we will have to start by dividing the amount he brings home by the gross.
156.4 / 184 = 0.85
By doing this, we get the answer in decimal form, which is not what we want.
Multiply it by 100 to convert it into a percent.
0.85 x 100 = 85.
Now, we know that he brings home 85% of his gross.