Answer: w=5 , L= 2+2(5)=12
Step-by-step explanation:
L=2+2W
A= 60
LxW=60, now we will replace the L
(2+2W)(W)=60 we multiply
2w^2+2w=60
2w^2+2w-60=0 we divide by 2 the equation so we can work easier
w^2+w-30=0
find out w using the quadratic equation
we will get 2 solution,
w=5 and another solution is -6, which is not valid, as the side can not be negative
a) The first integral corresponds to the area under y = f(x) on the interval [0, 3], which is a right triangle with base 3 and height 5, hence the integral is
b) The integral is zero since the areas under the curve over [3, 4] and [4, 5] are equal but opposite in sign. In other words, on the interval [3, 5], f(x) is symmetric and odd about x = 4, so
c) The integral over [5, 9] is the negative of the area of a rectangle with length 9 - 5 = 4 and height 5, so
Then by linearity, we have
Answers: b, d and e
b.The graph has a relative minimum
d. The graph has an x intercept at 3,0
e. the graph has an y intercept at 0,-15
f(x)=(x+5)(x-3)
The given equation is in the form of f(x) = a(x-b)(x-c)
If 'a' is positive then graph has a relative minimum
If 'a' is negative then graph has a relative maximum
Here a=1 that is positive so graph has a relative minimum .
To find x intercept we set f(x) =0 and solve for x
0=(x+5)(x-3)
x+5 =0 -> x = -5 so x intercept is (-5,0)
x - 3=0 -> x= 3 so x intercept is (3,0)
To find y intercept we plug in 0 for x
y=(x+5)(x-3)
y=(0+5)(0-3) = -15
so y intercept is (0,-15)
You have an algebraic expression in which you are solving for x. When you are solving for a variable, you need to isolate it all alone. Since you are multiplying by 1/5, you will have to undo it by multiplying by its reciprocal. In this case you are multiplying both sides by 5/1.
5/1 *1/5x = 121*5/1
x = 605
To check your answer, plug this value in for x and multiply it by 1/5. You should arrive at 121! Good luck!
