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
Given:
Side length of square = 2 in
To find:
The area of the square
Solution:
Area of the square:
Area = side × side
= 2 in × 2 in
Area = 4 in²
Therefore, the area of the square is 4 inches².
Answer:
Step-by-step explanation:
1) x - 21x = 5
x = x * 1
21x = x * 21
GCF = x
x - 21x = 5
x*(1 -21) = 5
x * (-20) = 5
-20x = 5
x =5/-20
x = -1/4
2) 6n² = 6 * n * n
30n = 6 * 5 * n
GCF = 6*n = 6n
6n² + 30n = 0
6n*(n + 5 ) = 0
6n = 0 or n + 5 = 0
n = 0 n = -5
n = 0 or -5
Answer:
5/6
Step-by-step explanation:
Rise over run
y2-y1 / x2-x1
30-15 / 36-18
Hope this helps!
