y = mx + b
y = 2x + 1
y - y₁ = m(x - x₁)
y - 4 = 2(x - 5)
y - 4 = 2(x) - 2(5)
y - 4 = 2x - 10
+ 4 + 4
y = 2x - 6
The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in
<h3>What is an
equation?</h3>
An equation is an expression that shows the relationship between two or more variables and numbers.
Let w represent the width, hence:
length = w + 33, height = w - 13
Volume (V) = w(w + 33)(w - 13) = w³ + 20w² - 429w
V(w) = w³ + 20w² - 429w
Rate of change = dV/dw = 3w² + 40w - 429
When w = 38, dV/dw = 3(38)² + 40(38) - 429 = 5423
When w = 53, dV/dw = 3(53)² + 40(53) - 429 = 10118
Rate = 10118 - 5423 = 4695 in³/in
The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in
Find out more on equation at: brainly.com/question/2972832
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F=5 dollar bills
t=twent dollar bills
38 total bills
f+t=38
400 in bills
400=5f+20t
we have
f+t=38
5f+20t=400
multipy first equation by -5 and add to second equation
-5f-5t=-190
5f+20t=400 +
0f+15t=210
15t=210
divide both sides by 15
t=14
subsitute
t+f=38
15+f=38
f=23
23 five dolar bills
15 twenty dollar bills
Answer:
the answer is 45 Lol.Hope that helps
9514 1404 393
Answer:
2,664,118.8 cm³/min
Step-by-step explanation:
The diameter at the water level of 5 m is (5/8)(6 m) = 3.75 m, so the area of the water surface is ...
A = (π/4)d² = (π/4)(3.75 m)² ≈ 11.044662 m²
When rising at the rate of 24 cm/min, the volume is increasing at the rate ...
(110,446.62 cm²)(24 cm/min = 2,650718.8 cm³/min
The input volume must be sufficient to accommodate this increase as well as the leakage. Then the input volume is ...
2,650718.8 cm³/min + 13,400 cm³/min = 2,664,118.8 cm³/min
