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Pani-rosa [81]
3 years ago
5

Which number would be rounded UP to the nearest ten but DOWN to the nearest hundred?

Mathematics
2 answers:
never [62]3 years ago
7 0

Answer:

B

Step-by-step explanation:

shtirl [24]3 years ago
3 0
B! i hope this helps
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I need help with these because recently we have been learning this and I don't get any of this.
Aleksandr-060686 [28]
It is upside down, can't see it properly!
5 0
3 years ago
CAN SOMEONE HELP ME WITH THIS QUESTION!!!
earnstyle [38]

Answer:

E

Step-by-step explanation:

let a, b, c represent the number of students in 6th, 7th, 8th grade

ratio of students : teachers = 28 : 1

There are 82 teachers , so 28 × 82 = 2296 students

Then

a + b + c = 2296 , that is

828 + b + c = 2296 ( subtract 828 from both sides )

b + c = 1468 → E

8 0
3 years ago
3y''-6y'+6y=e*x sexcx
Simora [160]
From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

3r^2-6r+6=0\iff r^2-2r+2=0

which has roots at r=1\pm i. This admits the two fundamental solutions

y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
7 0
3 years ago
Select the solution(s) of the original equation. 3+ √21/2+4 3+ √21/2-4 3- √21/2+4 3-√21/2-4​
jarptica [38.1K]

Answer:

2nd and 4th on edg

Step-by-step explanation:

3+21pi / 2 +4

3-21pi / 2 -4

4 0
2 years ago
Read 2 more answers
HELP PLEASE!!!
TiliK225 [7]

Problem 1

Answer: Independent

The reason why is because each bag is separate from one another, so one event doesn't affect the other. If we know the result of what we pulled out of one bag, it doesn't change the probability of the other event.

======================================

Problem 2

Answer: Dependent

Assuming you do not put the first card back, then the probability of picking a King on the second draw will be different than if you picked a King on the first draw. With all 52 cards in the deck, the probability of getting a king is 4/52 = 1/13. It changes to 4/51 after we picked out an ace for the first card (and didn't put that first card back).

6 0
3 years ago
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