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Natasha2012 [34]
3 years ago
8

W = A/L (solve for A) A = W +L A = W-L A = W/L A = LW

Mathematics
2 answers:
blondinia [14]3 years ago
8 0

Answer:

A = LW

Step-by-step explanation:

A/L =W

A =WL

therefore, A = LW

Hope it helped

kipiarov [429]3 years ago
6 0
A=wl, so the answer is A :)
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How many 30 degree angles are in a 150 degree angle? Use repeated subtraction to solve.
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3 years ago
A homogeneous rectangular lamina has constant area density ρ. Find the moment of inertia of the lamina about one corner
frozen [14]

Answer:

I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)

Step-by-step explanation:

By applying the concept of calculus;

the moment of inertia of the lamina about one corner I_{corner} is:

I_{corner} = \int\limits \int\limits_R (x^2+y^2)  \rho d A \\ \\ I_{corner} = \int\limits^a_0\int\limits^b_0 \rho(x^2+y^2) dy dx

where :

(a and b are the length and the breath of the rectangle respectively )

I_{corner} =  \rho \int\limits^a_0 {x^2y}+ \frac{y^3}{3} |^ {^ b}_{_0} \, dx

I_{corner} =  \rho \int\limits^a_0 (bx^2 + \frac{b^3}{3})dx

I_{corner} =  \rho [\frac{bx^3}{3}+ \frac{b^3x}{3}]^ {^ a} _{_0}

I_{corner} =  \rho [\frac{a^3b}{3}+ \frac{ab^3}{3}]

I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)

Thus; the moment of inertia of the lamina about one corner is I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)

7 0
3 years ago
(x+ delta x)^4 please help me
alexandr402 [8]
The awnser would be , 4x+4adetx
8 0
3 years ago
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amid [387]
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3 0
3 years ago
Read 2 more answers
With
Hatshy [7]
<span>y - 2418.6 = 5.9(x - 54)
y - 2418.6 = 5.9x - 318.6
y = 5.9x + 2100 
f(x) = 5.9x + 2100 (this is our function)
.
Now we can answer:
What is the weight of the plane with 26 gallons of fuel in its tank? 
f(26) = 5.9(26) + 2100 = 2253.4 pounds 
 (Put me as brainiest!!!! <3)</span>
3 0
3 years ago
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