i will solve it first i need full form
Answer:
2X/(3+X)
Step-by-step explanation:
X = 3m/2-m
Cross multiply both sides
X × (2-m)= 3m
2X-Xm= 3m
2X= 3m+Xm
2X= m(3+X)
Divide both sides by the coefficient of m which is (3+X)
m= 2X/(3+X)
The problem deals with fractions comparison, lets do it:
21/30 > 2/3
we begin solving:
21 > (2/3)*30
21 > 2*10
<span>21 > 20
</span>therefore the proposed inequality is true, <span>21/30 > 2/3
You can solve as well getting same denominator for both fractions and comparing directly, in this case we need to get 2/3 to be divided by 30:
2/3 = (10/10)(2/3) = 20/30
So we have:
</span><span>21/30 > 2/3
</span>which is equal to:
<span>21/30 > 20/30
</span>and we compare directly because both fractions are divided by the same number, and we can see that the inequality is true.
<h3>
Answer:</h3>
B) 6 hours
<h3>
Step-by-step explanation:</h3>
Let h represent the total trip time. Then h/2 is the time spent traveling at each speed. The distance covered is ...
... distance = speed · time
The sum of the distances in each mode is the total distance.
... 1020 = 55·h/2 +285·h/2
... 2040 = h·(55+285) . . . . multiply by 2
... 2040/340 = h = 6 . . . . . . . divide by the coefficient of h
The trip took a total of 6 hours.
Answer:
The term exponential is often used.
Step-by-step explanation:
The term exponential is used to represent changes in population over time. The idea of (positive) exponential is that the higher the number, the higher the growth. You can relate this with a population, because the higher the population, the more opportunities for it to multiply, thus, the higher it grows.
Usually the way to meassure the population of an species after certain number of years x, you use an exponential function of the form

For certain constants K₀ and a. K₀ is the initial population at the start of the experiment and <em>a </em>number of exponential growth. Essentially, the population of the species is multiplied by a during each year.
For example, if K₀ = 1000 and a = 2, then the population at the start of the experiment is 1000. After the first year is 1000*2 = 2000 and after the second year it is 2000*2 = 4000. Note that, not only the population grow during the years, but also the amount that the population increases also grow: in the first year it grows 1000, and between the first and second year it grows 2000.