Simplify equation: 14 - 7y + 5 = 1 + 3y + 24
simplify again: 19 - 7y = 25 + 3y
add -25 and 7y to both sides: -6 = 10y
y = -6/10 or -0.6
2. After 30 minutes, richie has delivered 1/3 of the papers, so they have to deliver the remaining 2/3. Since they can deliver them all in 40 minutes, would the answer be 2/3 of 40? I'm not sure about this.
Answer:
x+9
Step-by-step explanation:
if you expand...you could get x+9, which is the same as the given expression
Answer: 30 meters.
Step-by-step explanation:
Since the scale used by Bruce in the drawing that she made is:
1 milimeter = 10 meters
If the restaurant kitchen is 3 milimeters long in the drawing, the length of the actual kitchen will then be:
1/3 = 10/x
Cross multiply
1 × x = 3 × 10
x = 30
Therefore, the actual length is 30 meters.
<h3>Given</h3>
- a cone of height 0.4 m and diameter 0.3 m
- filling at the rate 0.004 m³/s
- fill height of 0.2 m at the time of interest
<h3>Find</h3>
- the rate of change of fill height at the time of interest
<h3>Solution</h3>
The cone is filled to half its depth at the time of interest, so the surface area of the filled portion will be (1/2)² times the surface area of the top of the cone. The filled portion has an area of
... A = (1/4)(π/4)d² = (π/16)(0.3 m)² = 0.09π/16 m²
This area multiplied by the rate of change of fill height (dh/dt) will give the rate of change of volume.
... (0.09π/16 m²)×dh/dt = dV/dt = 0.004 m³/s
Dividing by the coefficient of dh/dt, we get
... dh/dt = 0.004·16/(0.09π) m/s
... dh/dt = 32/(45π) m/s ≈ 0.22635 m/s
_____
You can also write an equation for the filled volume in terms of the filled height, then differentiate and solve for dh/dt. When you do, you find the relation between rates of change of height and area are as described above. We have taken a "shortcut" based on the knowledge gained from solving it this way. (No arithmetic operations are saved. We only avoid the process of taking the derivative.)
Note that the cone dimensions mean the radius is 3/8 of the height.
V = (1/3)πr²h = (1/3)π(3/8·h)²·h = 3π/64·h³
dV/dt = 9π/64·h²·dh/dt
.004 = 9π/64·0.2²·dh/dt . . . substitute the given values
dh/dt = .004·64/(.04·9·π) = 32/(45π)
Answer:
17
Step-by-step explanation:
165/10=16.5 17 because the extra 5 students need a teacher !
