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Alinara [238K]
3 years ago
8

Last Sunday, the average temperature was 8\%8%8, percent higher than the average temperature two Sundays ago. The average temper

ature two Sundays ago was TTT degrees Celsius. Which of the following expressions could represent the average temperature last Sunday? Choose 2 answers: Choose 2 answers:
Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
5 0

Answer: The two answers are 1.08<em>T</em> and (1 + \frac{8}{100} )<em>T</em>.

Step-by-step explanation:

Try multiplying 1.08 times any random number. If you use 5,  1.08 times 5 gives you 5.4.  This makes sense. Solve the other other expression as well and substitute <em>T</em><em> </em> with the SAME number that you used before (5). Hope that helps!

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Ipod: $89.00 Discount: 17% Tax: 5%
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Step-by-step explanation:

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5% Sales Tax Chart ($0.00 - $59.70)

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4 0
4 years ago
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PIT_PIT [208]

In the smaller triangle, let y be the length of the smallest leg. Then the smallest leg in the larger triangle is 14-y.

Within the scope of the smaller triangle, we have y such that

x^2+y^2=13^2\implies y=\sqrt{13^2-x^2}

Then within the larger the triangle, we would have

x^2+(14-y)^2=15^2\iff x^2+\left(14-\sqrt{13^2-x^2}\right)^2=15^2

Now we can solve for x:

x^2+14^2-28\sqrt{13^2-x^2}+(13^2-x^2)=15^2

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8 0
3 years ago
Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8
Kitty [74]
Substitute z=\ln x, so that

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)

Then the ODE becomes


x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0
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which has the characteristic equation r^2+1=0 with roots at r=\pm i. This means the characteristic solution for y(z) is

y_C(z)=C_1\cos z+C_2\sin z

and in terms of y(x), this is

y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)

From the given initial conditions, we find

y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1
y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8

so the particular solution to the IVP is

y(x)=\cos(\ln x)+8\sin(\ln x)
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