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Rashid [163]
3 years ago
8

The graph below shows the total cost for lunch, c, when Will and his friends buy a large salad to share and several slices of pi

zza, p.
For each additional slice of pizza that is purchased, by how much does the total cost of lunch increase?

$

Mathematics
2 answers:
Helga [31]3 years ago
5 0

Answer:

2.50 dollars.

Step-by-step explanation:

Given:

Will and his friends buy a large salad to share and several slices of pizza, p.

The graph given shows the total cost for lunch, c and slices of pizza,p.

p is marked horizontally, and c vertically.

Observing the points marked we find that for every increase of one slice of pizza, the cost i.e. c on vertical line increases by 0.50 times vertical grid.

Since each grid equals 5 dollars vertically, we find 1/2 grid increase is equal to increase of 1/2(5 ) =2.50 dollars.

Thus for increase of one slice of pizza, 2.50 dollars cost is increased.

mafiozo [28]3 years ago
5 0

Answer:

it 2.50

hope it helps

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Answer:

Step-by-step explanation:

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What is the constant of proportionality in the table below?
kenny6666 [7]

Answer:  \dfrac{2}{5}

Step-by-step explanation:

The constant of proportionality 'k' between two variables ( one independent and one dependent ) is given by :-

k=\dfrac{\text{Value of dependent variable}}{\text{Value of independent variable}}

In the given table, independent variable = Days

Dependent variable = Total Miles

From first row,  Days = 3

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k=\dfrac{\dfrac{6}{5}}{3}=\dfrac{6}{5\times3}=\dfrac{2}{5}

Hence, the constant of proportionality in the table is \dfrac{2}{5} .

3 0
3 years ago
Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A
Vlad [161]

Answer:

y(x)=sin(2x)+2cos(2x)

Step-by-step explanation:

y''+4y=0

This is a homogeneous linear equation. So, assume a solution will be proportional to:

e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda

Now, substitute y(x)=e^{\lambda x} into the differential equation:

\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0

Using the characteristic equation:

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Solving for \lambda:

\lambda =\pm2i

These roots give the next solutions:

y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}

Where c_1 and c_2 are arbitrary constants. Now, the general solution is the sum of the previous solutions:

y(x)=c_1 e^{2ix} +c_2 e^{-2ix}

Using Euler's identity:

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Redefine:

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Since these are arbitrary constants

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Finally, the solution is given by:

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