Let's take the number as x.
two-fifths of a number = 2x / 5
decreased by 3 = (2x/5) - 3
(2x/5) - 3
(2*25/5) - 3
(2*5) - 3
10 - 3
7
Answer: V = 5.22m/s
Step-by-step explanation:
Given that the mass of the particles is proportional to the fifth power of the speed. That is
M = k V^5
Where M = mass and V = speed
K = constant of proportionality
A certain river normally flows at a speed of 3 miles per hour
V = 3 mph
M = unknown
M = k × 3^5
M = 243K
K = M/243 ........(1)
What must its speed be in order to transport particles that are 16 times as massive as usual
M = 16M
Using same formula
I.e M = KV^5
16M = M/243 × V^5
M will cancel out
16 = V^5/243
V^5 = 3888
V = 5.22m/s
Answer:
Step-by-step explanation:
These exponential forms can be written into radicals very easily as long as you remember the rule: The denominator of the rational exponent serves as the index of the radical and the numerator serves as the exponent on the radicand. Let's look at a rational exponent. 3/4 4 would be the index on the radical (the number that sits in the little dip of the radical sign) and 3 is the power on the base. So
can be written in radical form as
![\sqrt[4]{x^3}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E3%7D)
Let's do 3^3/2 in your problem. 2 is the index (which is a "normal" square root and you don't need to write a 2 there cuz it's understood that it's a 2 if nothing is there), and 3 is the power on the base, which in our case is a 3. Bases can be numbers OR letters.
![3^{\frac{3}{2}}=\sqrt[2]{3^3}=\sqrt{3^3}](https://tex.z-dn.net/?f=3%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%3D%5Csqrt%5B2%5D%7B3%5E3%7D%3D%5Csqrt%7B3%5E3%7D)
That does in fact have an actual number answer, but I don't think you are simplifying them yet, only learning to write them from one form to another, so there you go!