1/2 and 10/20
to get 1/2 you have to divide 5/10 by 5:
5 / 5 1
----- = -----
10 / 5 2
to get 10/20 you have to multiply 5/10 by 2:
5 * 2 10
----- = ----
10 * 2 20
Hope it helps! =)
Answer:
a ) dAt/dt = 50,24 in/min
dh/dt = - 0,125 in/min
Step-by-step explanation:
The area of the top is At :
At = π*r²
a) Tacking derivatives with respect to time:
dAt/dt = 2* π*r * dr/dt
At t = t₁ r = 16 in and dr/dt = 0,5
Then
dAt/dt = 2*3,14*16*0,5 in/min
a ) dAt/dt = 50,24 in/min
b) The volume of the cylinder is:
Vc = π*r²*h ( where h is the heigh of the cylinder )
Tacking derivatives with respect to time
dVc/dt = 2* π*r*h*dr/dt + π*r²*dh/dt
But dVc/dt = 0 since the volume remains constant, then:
π*r²*dh/dt = - 2* π*r*h*dr/dt
r*dh/dt = - 2*h*dr/dt
dh/dt = - 2*0,5*2/16 in/min
dh/dt = - 0,125 in/min
<span> first, write the equation of the parabola in the required form: </span>
<span>(y - k) = a·(x - h)² </span>
<span>Here, (h, k) is given as (-1, -16). </span>
<span>So you have: </span>
<span>(y + 16) = a · (x + 1)² </span>
<span>Unfortunately, a is not given. However, you do know one additional point on the parabola: (0, -15): </span>
<span>-15 + 16 = a· (0 + 1)² </span>
<span>.·. a = 1 </span>
<span>.·. the equation of the parabola in vertex form is </span>
<span>y + 16 = (x + 1)² </span>
<span>The x-intercepts are the values of x that make y = 0. So, let y = 0: </span>
<span>0 + 16 = (x + 1)² </span>
<span>16 = (x + 1)² </span>
<span>We are trying to solve for x, so take the square root of both sides - but be CAREFUL! </span>
<span>± 4 = x + 1 ...... remember both the positive and negative roots of 16...... </span>
<span>Solving for x: </span>
<span>x = -1 + 4, x = -1 - 4 </span>
<span>x = 3, x = -5. </span>
<span>Or, if you prefer, (3, 0), (-5, 0). </span>
Given:
Volume of cuboid container = 2 litres
The container has a square base.
Its height is double the length of each edge on its base.
To find:
The height of the container.
Solution:
We know that,
1 litre = 1000 cubic cm
2 litre = 2000 cubic cm
Let x be the length of each edge on its base. Then the height of the container is:
The volume of a cuboid is:
Where, l is length, w is width and h is height.
Putting 
, we get
Divide both sides by 2.
Taking cube root on both sides.
![\sqrt[3]{1000}=x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1000%7D%3Dx)
Now, the height of the container is:
Therefore, the height of the container is 20 cm.
