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svetlana [45]
3 years ago
12

Karen gaines invested $15,000 in a money market account with an interest rate of 2.25% compounded semiannually. Five years later

, Karen withdrew the full amount to put towards the down payment on a new house. How much did Karen withdraw from the account
Mathematics
1 answer:
fredd [130]3 years ago
6 0

Answer:

years later , karen withdrew the full amount to put toward the down payment on a new house. How much did she withdraw? ... gaines invested $15,000 in a money market account with an interest rate of 2.75% compounded semiannually. ... Median response time is 34 minutes and may be longer for new subjects.

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Distribute the expression. 2(8nˆ2−5n+11)
Solnce55 [7]
2(8n^2-5n+11)
16n^2-5n+11
11n^2+11
6 0
3 years ago
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Built around 2600 BCE, the Great Pyramid of Giza in Egypt is 485 ft high (due to erosion, its current height is slightly less) a
Morgarella [4.7K]

Answer:1.69*10^12 J

Step-by-step explanation:

From figure above, using triangle ratio

485/755.5=y/l. Cross multiplying 485l=755.5y Divide via 485) hence l= 755.5y/485

Consider a slice volume Vslice= (755.5y/485)^2∆y; recall density =150lb/ft^3

Force slice = 150*755.5^2.y^2.∆y/485^2

From figure 2 in the attachment work done for elementary sclice

Wslice= 150.755.5^2.y^2.∆y.(485-y)/485^2

= (150*755.5^2*y^2)(485-y)∆y/485

To calculate the total work we integrate from y=0 to y= 485

Ie W=[ integral of 150*755.5^2 *y^2(485-y)dy/485] at y=0 and y= 485

Integrating the above

W= 150*755.5^2/485[485*y^3/3-y^4/4] at y= 0 and y=485

W= 150*755.5^2/485(485*485^3/3-484^4/4)-(485.0^3/3-0^4/4)

Work done 1.69*10^12joules

7 0
4 years ago
Drag each label to the correct location on the table. Each label can be used more than once, but not all labels will be used.
Natalka [10]

Answer:

what lable i dont see nothing

Step-by-step explanation:

7 0
3 years ago
Travelling only through the crust, surface waves arrive after body waves. though they arrive later than body waves, surface wave
leva [86]
I believe the answer would be volcanic eruptions.
4 0
3 years ago
Read 2 more answers
Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
I am Lyosha [343]
Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
7 0
3 years ago
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