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timurjin [86]
3 years ago
8

i need you to increase the number of customers you talk to daily by 20% i talk to an average of 8 customers per hour during an 8

hour shift so now i’ll need to talk to how many customers per day?

Mathematics
1 answer:
Scrat [10]3 years ago
4 0
I hope this helped your question

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How much larger is 6 times 10 to the 5th power compared to 2 times 10 to the 3rd power
lora16 [44]

Answer:

Either <em><u>10 times</u></em> or <u><em>598,000.</em></u>

Step-by-step explanation:

6 x 10 ^ 5 = 600,000

2 x 10 ^ 3 = 2,000

If we are figuring out the exact number, 600,000 - 2,000.  If we are finding out how many powers larger, count.

600,000 - 2,000 = 598,000

600,000 is 10 times larger than 2,000.

See?

600,0<u>00</u>

2,000

6 0
3 years ago
Two boats start their journey from the same point A and travel long directions AC and AD as shown below what is the distance CD
Olegator [25]

I think its B but ima have to check again and get back to you

4 0
3 years ago
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Ten times the sum of half a number and 6 is 8
Gelneren [198K]
10(1/2x + 6) = 8
5x + 60 = 8
5x = 8 - 60
5x = - 52
x = -52/5
5 0
3 years ago
Read 2 more answers
A supermarket has a length of 150 feet.the area of the supermarket is 1,350 feet squared.what is the width of the supermarket
Amiraneli [1.4K]

Answer:

1,500

Step-by-step explanation:

8 0
3 years ago
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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