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laila [671]
3 years ago
11

3. What is 5⅕% of 325?

Mathematics
1 answer:
Naya [18.7K]3 years ago
3 0

Answer:

20% × 325 =

(20 ÷ 100) × 325 =

(20 × 325) ÷ 100 =

6,500 ÷ 100 =

65

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Idk can someone help
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Answer:

if its 2/7 then 98 the luck also counts so 20

Step-by-step explanation:

luck+98+the premtier

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In a fruit punch drink, the 3 ingredients are apple juice, orange juice and cranberry juice. If 3/4of the drink is apple juice a
Vilka [71]
7:1:2 7 is apple juice 1 is orange juice and 2 is cranberry juice
2:7:1 would be the final answer
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Rebecca drove 44 1/3 miles in 1 1/3 hours. What is the unit rate for miles per hour?
saul85 [17]
unit\ rate=\frac{distance}{time}\\\\
distance=44\frac{1}{3} miles\\
time=1\frac{1}{3}hour\\\\
unit\ rate=\frac{44\frac{1}{3}}{1\frac{1}{3}}=\ \ \ | make\ improper\ fractions\\
\frac{\frac{133}{3}}{\frac{4}{3}}=\frac{133}{3}*\frac{3}{4}=\frac{133}{4}=33\frac{1}{4}\\\\
unit\ rate= 33\frac{1}{4}\frac{miles}{hour}

5 0
3 years ago
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
Cary sets up a checking account with an initial balance of $27,700, and the rent for her
jek_recluse [69]

Answer:

$1,650

Step-by-step explanation:

First, we need to find how much she paid the whole year. 27,700-x=7900. If we subtract 7900 from each side and add x to each side, we get 19,800=x. Therefore, she paid $19,800 in rent the whole year. Since there are 12 months in a year, and each month she paid the same, she paid $1,650 in rent each year, for 19800/12=1650

5 0
3 years ago
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