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3 years ago

Clayton is at most 2 meters above sea level. Using C for Clayton, which inequality correctly represents this statement?

Mathematics

1 answer:

Nataly [62]3 years ago

5 0

Answer:

C ≤ 2

Step-by-step explanation:

Given that :

Clayton is at most 2 meters above sea level ;

Let Clayton = C

Clayton's distance can be represented by the inequality :

2 meters above sea level is positive, +2

At most 2 meters above sea level means, Clayton could be anywhere between + 2 metwrs but not more than 2 meters

Hence, Clayton's distance can be represented by the inequality :

C ≤ 2

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There are nine water bottles in a fridge. Three full boxes are added. Two more boxes are added, both of which have one fewer bot

Degger [83]

The number of bottles in a full box is 12.

Explanation

Suppose, the number of bottles in a full box is $x$

So, the number of bottles in three full boxes $=3x$

Now, two more boxes are added, both of which have one fewer bottle than the other three. So, the number of bottles in these two boxes $= 2(x-1)$

As there were 9 bottles initially and now there are 67 bottles in the fridge, so the equation will be....

$9+3x+2(x-1)=67\\ \\ 9+3x+2x-2=67\\ \\ 5x+7=67\\ \\ 5x=67-7\\ \\ 5x=60\\ \\ x=\frac{60}{5}=12$

Thus, the number of bottles in a full box is 12.

7 0

3 years ago

Do not need a explanation but would be nice!

djyliett [7]

Answer: 5-35i

Step-by-step explanation:

I have no explanation. I'm sorry.

8 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:

Point a:

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.

Point b:

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 y we get:

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

Point c:

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

What is the volume of a cylinder that has the following measurements? area of each base: 50.24 ft2 area of lateral surface: 75.3

Andrew [12]

Answer:

150.72 ft³

Step-by-step explanation:

Given:

Area of the base :

B = 50.24 ft²

Height of cylinder:

h = 3 ft

The volume is:

V = Bh
V = 50.24*3 = 150.72 ft³

5 0

3 years ago

Read 2 more answers

I need help with this math problem.

user100 [1]

<h3>Answer: Choice B) </h3><h3>-6x - 2y = 12</h3>

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

Explanation:

The x intercept is (-2,0) which is where the graph crosses the x axis.

The y intercept is (0,-6) which is where the graph crosses the y axis.

-----

Find the slope of the line through those two points

m = (y2-y1)/(x2-x1)

m = (-6-0)/(0-(-2))

m = (-6-0)/(0+2)

m = -6/2

m = -3

-----

The y intercept (0,-6) leads to b = -6

Both m = -3 and b = -6 plug into y = mx+b to get

y = mx+b

y = -3x+(-6)

y = -3x-6

-----

Now add 3x to both sides

y = -3x-6

y+3x = -3x-6+3x

3x+y = -6

-----

Lastly, multiply both sides by -2 so that the "-6" on the right hand side turns into "12" (each answer choice has 12 on the right hand side)

3x+y = -6

-2(3x+y) = -2(-6)

-2(3x)-2(y) = 12

-6x-2y = 12

which is what choice B shows.

5 0

3 years ago