Answer:
ok thanks for ur care my friend have a wonderful day
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
well, it's isosceles so use base angles theorem
the top angle is also x
90 + 2x = 180
subtract 90 from both sides
2x = 90
divide both sides by 2
x = 45 degrees
7. i would guess 95-100 people were surveyed. the way you estimate this is by adding the number of students in each category. ex. about 3 in the first bar, about 9 in the next
8. no, you could not list the exact height because the data given only shows the number of students in between 75-80, not their actual height.
Answer:
r=2
Step-by-step explanation:
r + 8 r + 11 = 29
( 1 + 8 ) r + 11 = 29 9 r + 11 = 29
Now we can isolate and solve for r while always keeping the equation balanced: First, subtract 11 from each side of the equation:
9 r + 11 − 11 = 29 − 11
9 r + 0 = 18
9 r = 18
Now we can divide each side of the equation by 9 to get
r : 9 over 9 = 18 under 9
1 r = 2
r = 2
