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olchik [2.2K]
3 years ago
13

At D.J.s Drink stand Erika ordered a cup of fruit punch made using 1/4 cup of lemonade 1/12 cup of cranberry juice. What fractio

n will be orange juice?
Mathematics
1 answer:
Zepler [3.9K]3 years ago
8 0

The fraction of the orange juice will be 2/3cup.

Why?

Since there is no additional information, let's assume that the fruit punch is made using just three juice fruits, lemonade juice, cranberry juice and the rest is orange juice.

We can calculate the fraction of the orange juice using the following information:

FruitPunch=\frac{1}{4}Lemonade+\frac{1}{12}Cranberry+Orange\\\\1cup=\frac{1}{4}cup+\frac{1}{12}cup+Orange\\\\1cup-(\frac{1}{4}cup+\frac{1}{12}cup)=Orange\\\\1cup-(\frac{12+4}{4*12})=Orange\\\\Orange=1cup-\frac{16}{48}cup=1cup-\frac{1}{3}cup=\frac{3-1}{3}cup=\frac{2}{3}cup

Hence, we have that the fraction of the orange juice will be 2/3cup.

Have a nice day!

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The domain is all real numbers, reason being, you can put any number for x and get a valid coordinate.
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A cell phone tower covers a circular area with a radius of 15 miles. You are at a location that is 11 miles west and 12 miles so
Murljashka [212]

Answer:

No you do not.

This is because we can use Pythagoras Theorem to show that our location sits outside the 15 mile radius of the cell tower.

To work is out you would write the equation

{11}^{2}  +  {12}^{2}  = 265

Then to work out the Hypotheses you would

\sqrt{265}  = 16.2788205961

Which proves that the location is outside of the 15 Mile radius of the cell tower.

3 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
1.) What is the product of (-6)^2?
Dmitriy789 [7]

im not sure about the first one but im 100% about 2 and 3

2-A

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5 0
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Y = –8x + 8<br> y = –8x + 8
fiasKO [112]

Answer:

x=0

y=8

Step-by-step explanation:

This is a simultaneous equation

So let's solve

y=-8x+8..(1)

y=-8x+8...(2)

Add (1) and (2)

2y=16

Make y the subject of formula by dividing both sides by 2

y=8

Substitute the value for y into (2)

8=-8x+8

Collect like terms

8-8=-8x

0=-8x

Divide both sides by-8

x=0

Therefore x is 0 and y is 8

5 0
3 years ago
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