When a number is 3 times a first number. A third number is 100 more than the first number and the sum of the three numbers is 450, the numbers are 70, 170 and 210.
<h3>How to calculate the numbers?</h3>
Let the first number = x
Second number = 3x
Third number = x + 100
Sum = 450
The numbers will be:
x + 3x + x + 100 = 450
5x + 100 = 450
5x = 450 - 100
5x = 350
Divide
x = 350 / 5
x = 70
First number = 70
Second number = 3x = 3 × 70 = 210
Third number = x + 100 = 70 + 100 = 170
Therefore, the numbers are 70, 170 and 210.
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One number is 3 times a first number. A third number is 100 more than the first number. If the sum of the three numbers is 450 , find the numbers.
The solution is the interval b is in ]11.3, infinity[
Answer:
12=8 18=12 24=16
Step-by-step explanation:
You time the 4 by the amount of times the 6 is timed.
Answer:
X^2-18x+81
OR
(X-9)^2
Step-by-step explanation:
X(x-9)-9(x-9)=
x^2-9x-9x+81
X^2-18x+81
The table is attached in the figure.
g(x) = f(4x) ⇒⇒⇒ differentiating both sides with respect to x
∴ g'(x) = ![\frac{d}{dx} [f(x)] * \frac{d}{dx} [4x]=4*f'(x)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bf%28x%29%5D%20%2A%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5B4x%5D%3D4%2Af%27%28x%29)
⇒⇒⇒⇒⇒⇒ chain role
To find g '(0.1)
Substitute with x = 0.1
from table:
f'(0.1) = 1 ⇒ from the table
∴ g'(0.1) = 4 * [ f'(0.1) ] = 4 * 1 = 4
