To get the solution, we are looking for, we need to point out what we know.
1. We assume, that the number 112 is 100% - because it's the output value of the task.
2. We assume, that x is the value we are looking for.
3. If 112 is 100%, so we can write it down as 112=100%.
4. We know, that x is 200% of the output value, so we can write it down as x=200%.
5. Now we have two simple equations:
1) 112=100%
2) x=200%
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
112/x=100%/200%
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.
7. Solution for what is 200% of 112
112/x=100/200
(112/x)*x=(100/200)*x - we multiply both sides of the equation by x
112=0.5*x - we divide both sides of the equation by (0.5) to get x
112/0.5=x
224=x
x=224
now we have:
200% of 112=224
●I tried my best I was never good with percentages my least favorite.... Please let me know if you got it wrong. If you do I'm sorry.
∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
----
∫(t = 2 to 3) t √(t - 2) dt
= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2
= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du
= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}
= 26/15.
----
For the k-entry, use integration by parts with
u = t, dv = sin(πt) dt
du = 1 dt, v = (-1/π) cos(πt).
So, ∫(t = 2 to 3) t sin(πt) dt
= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt
= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]
= 5/π + 0
= 5/π.
Therefore,
∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.
Question 21
Selling price = 18 + 18 x 36/100
Question 22
Discount price = 80 - 80x10/100
Selling price = discount price + discount price x 7/100
Answer:
x = -3
x = -1
x = 2
Step-by-step explanation:
The <u>zeros of a function</u> are the x-values of the points at which the curve crosses the x-axis.
From inspection of the given graph, the curve crosses the x-axis at:
Therefore, these are the zeros of the function.
