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kari74 [83]
1 year ago
7

1. Ms. Geyer wants to take personal training classes at a nearby gym but needs to start by selecting a membership plan.

Mathematics
1 answer:
Triss [41]1 year ago
5 0
Mrs geyer is gonna have to get a new membership to the gym which is 60$ of ever day
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Fill in the blank _+8=+6<br> And chose the property of addition you send
Paha777 [63]

Answer:

-2

Step-by-step explanation:

Just use 6-8 and you will get - 2

Check again with - 2+8 you will get positive 6

4 0
3 years ago
Which values of c will cause the quadratic equation –x2 + 3x + c = 0 to have no real number solutions? Check all that apply.
nika2105 [10]
Quadratic equation which has no solution gives zero for the value of d.
d = b² - 4ac
0 = 3² +- 4(-1)c
-9 = +_ 4c
c = -9/-4 pr 9/-4
c = 9/4 or -9/4

In short, Your Answers would be Option C & E

Hope this helps!


7 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
If logarithm of 5832 be 6, Find thw base?​
Lunna [17]

Answer:

3sqrt(2) ................

6 0
2 years ago
The shape below is transformed to another shape.
netineya [11]

Answer:

Option 4 is the image of the given figure.

Step-by-step explanation:

We are given that,

The shape EFGHCD is transformed to form another shape.

From the options, we see that,

Figure 2 and 3 does not have the same vertices as that of the figure.

So, they are discarded.

Since, after transforming a figure, we get a new figure.

So, the vertices cannot have same name as that of the original figure.

So, option 1 is discarded.

Thus, we get,

Option 4 is the image of the given figure after transformation as shown below.

5 0
3 years ago
Read 2 more answers
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