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3 years ago

Cost of a dozen eggs depends on the size of the eggs. what is the dependent variable?

2 answers:

6 0

Answer: the dependent variable is the cost of a dozen of eggs.

Explanation:

The cost is function of the size of the eggs, Cost = f (size).

That means that you introduce the size (this is the input) in the function an you obtain the cost (this is the output)

The input is the independent variable and the output is the dependent variable.

sineoko [7]3 years ago

4 0

the cost of the eggs
cost (x) = x*price
this is an example of a proportional cost relation with the size been x

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Lukas graphed the system of equations shown.

vampirchik [111]

The system of equations 2x + 3y = 2 and y = (1/2)x + 3 have solutions at x = -2 and y = 2

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

From the system of equations 2x + 3y = 2 and y = (1/2)x + 3, the graph of the equation shows that the solution is at x = -2 and y = 2

Find out more on equation at: brainly.com/question/2972832

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1 year ago

A grocery store sells a 7 oz bag of raisins for $1.10 and a 9 oz bag of raisins for $1.46. which size bag has the lower price pe

Marysya12 [62]

7 oz bag:
110 ÷ 7 = 15.7143¢ per oz
9 oz bag:
146 ÷ 9 = 16.2222¢ per oz
So the answer is the 7 oz bag, hope this helps!

5 0

3 years ago

Read 2 more answers

The point-slope form of the equation of a line that passes through points (8, 4) and (0, 2) is

Sladkaya [172]

Answer:

y= x-4

Step-by-step explanation:

5 0

3 years ago

Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

$y_p=a\sin x+b\cos x$

${y_p}'=a\cos x-b\sin x$

${y_p}''=-a\sin x-b\cos x$

Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

$y''+y=x\cos(2x)$

$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

$-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)$

$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

7 0

2 years ago

Nu is designing a rectangular sandbox.The bottomm is 16 square feet.Which dimensions require the least amount of materail for th

zzz [600]

Given Information:

Area of rectangle = 16 square feet

Required Information:

Least amount of material = ?

Answer:

x = 4 ft and y = 4 ft

Step-by-step explanation:

We know that a rectangle has area = xy and perimeter = 2x + 2y

We want to use least amount of material to design the sandbox which means we want to minimize the perimeter which can be done by taking the derivative of perimeter and then setting it equal to 0.

So we have

xy = 16

y = 16/x

p = 2x + 2y

put the value of y into the equation of perimeter

p = 2x + 2(16/x)

p = 2x + 32/x

Take derivative with respect to x

d/dt (2x + 32/x)

2 - 32/x²

set the derivative equal to zero to minimize the perimeter

2 - 32/x² = 0

32/x² = 2

x² = 32/2

x² = 16

x = $\sqrt{16} = 4$ ft

put the value of x into equation xy = 16

(4)y = 16

y = 16/4

y = 4 ft

So the dimensions are x = 4 ft and y = 4 ft in order to use least amount of material.

Verification:

xy = 16

4*4 = 16

16 = 16 (satisfied)

7 0

3 years ago