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11Alexandr11 [23.1K]
3 years ago
8

925 round to the nearest hundred

Mathematics
2 answers:
yarga [219]3 years ago
4 0
925 rounded to the nearest hundred is 900 becuase first we identify the hundreds digit which in this case is 9. Second,we identify the next smallest place value (the digit to the right of the hundreds place) which in this case is 2. Is that digit greater than or equal to five? No - we round Down. The hundreds digit is stays the same but every digit after it becomes a zero.
vesna_86 [32]3 years ago
3 0
925 round to the nearest hundred

2 is after or before 5?
Before so we will round back, which would make the answer 900. 

925 rounded to the nearest hundred  = 900<span />
You might be interested in
Let y(t) be the solution to y˙=3te−y satisfying y(0)=3 . (a) Use Euler's Method with time step h=0.2 to approximate y(0.2),y(0.4
OLEGan [10]

Answer:

  • y(0.2)=3, y(0.4)=3.005974448, y(0.6)=3.017852169, y(0.8)=3.035458382, and y(1.0)=3.058523645
  • The general solution is y=\ln \left(\frac{3t^2}{2}+e^3\right)
  • The error in the approximations to y(0.2), y(0.6), and y(1):

|y(0.2)-y_{1}|=0.002982771

|y(0.6)-y_{3}|=0.008677796

|y(1)-y_{5}|=0.013499859

Step-by-step explanation:

<em>Point a:</em>

The Euler's method states that:

y_{n+1}=y_n+h \cdot f \left(t_n, y_n \right) where t_{n+1}=t_n + h

We have that h=0.2, t_{0}=0, y_{0} =3, f(t,y)=3te^{-y}

  • We need to find y(0.2) for y'=3te^{-y}, when y(0)=3, h=0.2 using the Euler's method.

So you need to:

t_{1}=t_{0}+h=0+0.2=0.2

y\left(t_{1}\right)=y\left(0.2)=y_{1}=y_{0}+h \cdot f \left(t_{0}, y_{0} \right)=3+h \cdot f \left(0, 3 \right)=

=3 + 0.2 \cdot \left(0 \right)= 3

y(0.2)=3

  • We need to find y(0.4) for y'=3te^{-y}, when y(0)=3, h=0.2 using the Euler's method.

So you need to:

t_{2}=t_{1}+h=0.2+0.2=0.4

y\left(t_{2}\right)=y\left(0.4)=y_{2}=y_{1}+h \cdot f \left(t_{1}, y_{1} \right)=3+h \cdot f \left(0.2, 3 \right)=

=3 + 0.2 \cdot \left(0.02987224102)= 3.005974448

y(0.4)=3.005974448

The Euler's Method is detailed in the following table.

<em>Point b:</em>

To find the general solution of y'=3te^{-y} you need to:

Rewrite in the form of a first order separable ODE:

e^yy'\:=3t\\e^y\cdot \frac{dy}{dt} =3t\\e^y \:dy\:=3t\:dt

Integrate each side:

\int \:e^ydy=e^y+C

\int \:3t\:dt=\frac{3t^2}{2}+C

e^y+C=\frac{3t^2}{2}+C\\e^y=\frac{3t^2}{2}+C_{1}

We know the initial condition y(0) = 3, we are going to use it to find the value of C_{1}

e^3=\frac{3\left(0\right)^2}{2}+C_1\\C_1=e^3

So we have:

e^y=\frac{3t^2}{2}+e^3

Solving for <em>y</em> we get:

\ln \left(e^y\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y\ln \left(e\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y=\ln \left(\frac{3t^2}{2}+e^3\right)

<em>Point c:</em>

To compute the error in the approximations y(0.2), y(0.6), and y(1) you need to:

Find the values y(0.2), y(0.6), and y(1) using y=\ln \left(\frac{3t^2}{2}+e^3\right)

y(0.2)=\ln \left(\frac{3(0.2)^2}{2}+e^3\right)=3.002982771

y(0.6)=\ln \left(\frac{3(0.6)^2}{2}+e^3\right)=3.026529965

y(1)=\ln \left(\frac{3(1)^2}{2}+e^3\right)=3.072023504

Next, where y_{1}, y_{3}, \:and \:y_{5} are from the table.

|y(0.2)-y_{1}|=|3.002982771-3|=0.002982771

|y(0.6)-y_{3}|=|3.026529965-3.017852169|=0.008677796

|y(1)-y_{5}|=|3.072023504-3.058523645|=0.013499859

3 0
3 years ago
I don't know what this means someone help?????
liubo4ka [24]

Answer:

2

Step-by-step explanation:

so 10÷2=5 6÷2=3 3/5

that what u need?

3 0
3 years ago
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If Mary has 12 more mangoes than Jason and Jason has y mangoes how many has Mary?​
xeze [42]

Answer:

y+12=m

Step-by-step explanation:

m (marys mangos)

3 0
2 years ago
Round to the given place: 346,535 (ten thousands)
Paul [167]
350,000 because the ten thousands is 40,000 so you would look to the right of the four and it’s a six so you would add a one to the four to make it a five and replace all other numbers behind the five with zeros
6 0
3 years ago
Read 2 more answers
I need help !!!!!!!!!
solniwko [45]

Answer:

i think b

Step-by-step explanation:

3 0
3 years ago
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