Answer:
9 months.
Step-by-step explanation:
use a calculator
2 | <u>5</u><u>5</u><u>0</u><u>,</u><u>7</u><u>5</u><u>0</u><u>,</u><u>9</u><u>0</u><u>0</u>
2 | <u>2</u><u>7</u><u>5</u><u>,</u><u>3</u><u>7</u><u>5</u><u>,</u><u>4</u><u>5</u><u>0</u>
3 |<u> </u><u>2</u><u>7</u><u>5</u><u>,</u><u>3</u><u>7</u><u>5</u><u>,</u><u>2</u><u>2</u><u>5</u>
<u>3</u><u> </u>| <u>2</u><u>7</u><u>5</u><u>,</u><u>1</u><u>2</u><u>5</u><u>,</u><u>7</u><u>5</u>
5 | <u>2</u><u>7</u><u>5</u><u>,</u><u>1</u><u>2</u><u>5</u><u>,</u><u>2</u><u>5</u>
5 | <u>55,25,5</u>
5 | <u>1</u><u>1</u><u>,</u><u>5</u><u>,</u><u>5</u>
11 | <u>1</u><u>1</u><u>,</u><u>1</u><u>,</u><u>1</u>
LCM:-2×2×3×3×5×5×5×11=49500
Answer:
The x-coordinate of the point changing at ¼cm/s
Step-by-step explanation:
Given
y = √(3 + x³)
Point (1,2)
Increment Rate = dy/dt = 3cm/s
To calculate how fast is the x-coordinate of the point changing at that instant?
First, we calculate dy/dx
if y = √(3 + x³)
dy/dx = 3x²/(2√(3 + x³))
At (x,y) = (1,2)
dy/dx = 3(1)²/(2√(3 + 1³))
dy/dx = 3/2√4
dy/dx = 3/(2*2)
dy/dx = ¾
Then we calculate dx/dt
dx/dt = dy/dt ÷ dy/dx
Where dy/dx = ¾ and dy/dt = 3
dx/dt = ¾ ÷ 3
dx/dt = ¾ * ⅓
dx/dt = ¼cm/s
The x-coordinate of the point changing at ¼cm/s
<span>The coterminal angles of angle A are given by adding k times 360° to it, where k is an integer, positive or negative.
Similarly, all angles that are coterminal with 141° are given by
141° + k * 360° where k is a positive or negative integer.
That is 141° + 360°, 141 + 2*360° , 141+3*360° and so on
and 141° - 360°, 141° -2*360°, 141° - 3*360° and so on.
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