Hi,
Following the rules of PEMDAS you first subtract in the parentheses.
(5-2)=3
Then multiply the parentheses.
(3)(-6)=-18
Solve the exponent.
-4^2=16
Then you add -18 and 16 to get -2.
So the answer is B.-2.
I hope this helps.
Answer:
7.37108cmx7.37108cmx7.37108cm
Step-by-step explanation:
Find the volume of the cylinder then take the cube route of that. You should end up with ~7.38108cm which is the length, width, and height of your cube. The volume is 402.12386cm cubed.
The answer is a.0.32 km.
The speed that a tsunami can travel is modeled by the equation is s = 356√d.
It is given:
s = 200 km/h
d = ?
Now, let's substitute s in the equation and find d:
s = 356√d
200 = 356√d
√d = 200 ÷ 356
√d = 0.562
Now, let's square both sides of the equation:
(√d)² = (0.562)²
d = (0.562)² = 0.316 ≈ 0.32
Therefore, <span> the approximate depth (d) of water for a tsunami traveling at 200 kilometers per hour is 0.32 km.</span>
C. 37
It is the lower quartile marked on the box-and-whisker plot and is the median of all data below the median
