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Elden [556K]
2 years ago
12

A man has 10 more dimes than quarters. he has $2.75 how many of each coin does he have?

Mathematics
1 answer:
Anton [14]2 years ago
4 0
5 quarters, 15 dimes

q + 10 = d
0.25q + 0.1d = 2.75

0.25q + 0.1(q + 10) = 2.75
0.25q + 0.1q + 1 = 2.75
0.35q = 1.75
q = 5

Check:

0.25(5)+ 0.1(15)= 2.75
1.25 + 1.5 = 2.75
2.75 = 2.75 :)
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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)

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2 years ago
What is 5 minus 16 tenths
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The answer would be 3.4

Hope that helped : )
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Please helpppp asappp
viva [34]

Answer:

=561.2 square feet

Step-by-step explanation:

top rectangle=19×8=152

bottom rectangle=5×8=40

two slanting rectangles=8×14×2=224

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1/2×12×12.1×2=145.2

add them all

152+40+224+145.2=561.2

3 0
3 years ago
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