Answer:
<em>2 solutions</em>
Step-by-step explanation:
Given the expression
2m/2m+3 - 2m/2m-3 = 1
Find the LCM of the expression at the left hand side:
2m(2m-3)-2m(2m+3)/(2m+3)(2m-3) = 1
open the bracket
4m²-6m-4m²-6m/(4m²-9) = 1
Cross multiply
4m²-6m-4m²-6m = 4m² - 9
-12m = 4m² - 9
4m² - 9+12m = 0
4m² +12m-9 = 0
<em>Since the resulting equation is a quadratic equation, it will have 2 solutions since the degree of the equation is 2</em>
Just substitute each number into the expression:
5(5) - 6(3) + 20(1/4)/4(3)(1/4) = 25 - 18 + 5/3 = 7 + 5/3 = 26/3 or 8 and 2/3.
Answer:they made 14.76
Step-by-step explanation:
at the bake sale
Answer:
Using either method, we obtain: 
Step-by-step explanation:
a) By evaluating the integral:
![\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cint%5Climits%5Et_0%20%7B%5Csqrt%5B8%5D%7Bu%5E3%7D%20%7D%20%5C%2C%20du)
The integral itself can be evaluated by writing the root and exponent of the variable u as: ![\sqrt[8]{u^3} =u^{\frac{3}{8}](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7Bu%5E3%7D%20%3Du%5E%7B%5Cfrac%7B3%7D%7B8%7D)
Then, an antiderivative of this is: 
which evaluated between the limits of integration gives:
and now the derivative of this expression with respect to "t" is:
b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:
"If f is continuous on [a,b] then
is continuous on [a,b], differentiable on (a,b) and 
Since this this function 
is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:
