Answer:
(h+4)
Step-by-step explanation:
Answer:
5 x^2 y (9 x y + 1)
Step-by-step explanation:
Factor the following:
45 x^3 y^2 + 5 x^2 y
Factor common terms out of 45 x^3 y^2 + 5 x^2 y.
Factor 5 x^2 y out of 45 x^3 y^2 + 5 x^2 y:
Answer: 5 x^2 y (9 x y + 1)
Answer:
The system of linear equations are
and 
Step-by-step explanation:
Given : The number of students who chose lunch was 5 more than the number of students who chose breakfast. Let x represent the number of students who chose breakfast and y represent the number of students who chose lunch.
(50 students picked, 25 picked dinner the rest picked lunch and breakfast)
To find : Write a system of linear equations that represents the numbers of students who chose breakfast and lunch ?
Solution :
The number of students who chose breakfast be 'x'
The number of students who chose lunch be 'y'.
The number of students who chose lunch was 5 more than the number of students who chose breakfast.
i.e. 
Now, Total student were 50 and 25 picked dinner the rest picked lunch and breakfast i.e. 25.
So, 
Therefore, the system of linear equations are
and 
Answer:
-2, 8/3
Step-by-step explanation:
You can consider the area to be that of a trapezoid with parallel bases f(a) and f(4), and width (4-a). The area of that trapezoid is ...
A = (1/2)(f(a) +f(4))(4 -a)
= (1/2)((3a -1) +(3·4 -1))(4 -a)
= (1/2)(3a +10)(4 -a)
We want this area to be 12, so we can substitute that value for A and solve for "a".
12 = (1/2)(3a +10)(4 -a)
24 = (3a +10)(4 -a) = -3a² +2a +40
3a² -2a -16 = 0 . . . . . . subtract the right side
(3a -8)(a +2) = 0 . . . . . factor
Values of "a" that make these factors zero are ...
a = 8/3, a = -2
The values of "a" that make the area under the curve equal to 12 are -2 and 8/3.
_____
<em>Alternate solution</em>
The attachment shows a solution using the numerical integration function of a graphing calculator. The area under the curve of function f(x) on the interval [a, 4] is the integral of f(x) on that interval. Perhaps confusingly, we have called that area f(a). As we have seen above, the area is a quadratic function of "a". I find it convenient to use a calculator's functions to solve problems like this where possible.