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liq [111]
3 years ago
8

A cheese pizza costs $5. Additional toppings cost $1.50 each. Write and graph an equation that represents the total cost $c$ (in

dollars) of a pizza with $t$ toppings.
An equation that represents the total cost of a pizza c=____
Mathematics
2 answers:
anzhelika [568]3 years ago
8 0

Answer:

1.5x + 5 = total cost

Step-by-step explanation:

If you have x toppings you multiply 1.5 by x. :)

Triss [41]3 years ago
7 0

Answer:

5 + 1.5x = total cost

Step-by-step explanation:

5 is for one pizza and x is for toppings. For every topping it is multiplied by 1.5. for example, if you have 2 toppings for one cheese pizza it would be 5 dollars + 1.5x2 which is 3 so 3+5=8$ in total.

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Please help with this question.
Kisachek [45]
The answer is the top option.
5 0
3 years ago
Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x​
Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

<em>r</em> ² - 3<em>r</em> + 2 = (<em>r</em> - 2) (<em>r</em> - 1) = 0

with roots <em>r</em> = 2 and <em>r</em> = 1, so the characteristic solution is

<em>y</em> (char.) = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>)

For the <em>ansatz</em> particular solution, we might first try

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> + <em>d</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

where <em>ax</em> + <em>b</em> corresponds to the 2<em>x</em> term on the right side, (<em>cx</em> + <em>d</em>) exp(<em>x</em>) corresponds to (1 + 2<em>x</em>) exp(<em>x</em>), and <em>e</em> exp(3<em>x</em>) corresponds to 4 exp(3<em>x</em>).

However, exp(<em>x</em>) is already accounted for in the characteristic solution, we multiply the second group by <em>x</em> :

<em>y</em> (part.) = (<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)

Now take the derivatives of <em>y</em> (part.), substitute them into the DE, and solve for the coefficients.

<em>y'</em> (part.) = <em>a</em> + (2<em>cx</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

… = <em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)

<em>y''</em> (part.) = (2<em>cx</em> + 2<em>c</em> + <em>d</em>) exp(<em>x</em>) + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… = (<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

Substituting every relevant expression and simplifying reduces the equation to

(<em>cx</em> ² + (4<em>c</em> + <em>d</em>)<em>x</em> + 2<em>c</em> + 2<em>d</em>) exp(<em>x</em>) + 9<em>e</em> exp(3<em>x</em>)

… - 3 [<em>a</em> + (<em>cx</em> ² + (2<em>c</em> + <em>d</em>)<em>x</em> + <em>d</em>) exp(<em>x</em>) + 3<em>e</em> exp(3<em>x</em>)]

… +2 [(<em>ax</em> + <em>b</em>) + (<em>cx</em> ² + <em>dx</em>) exp(<em>x</em>) + <em>e</em> exp(3<em>x</em>)]

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

… … …

2<em>ax</em> - 3<em>a</em> + 2<em>b</em> + (-2<em>cx</em> + 2<em>c</em> - <em>d</em>) exp(<em>x</em>) + 2<em>e</em> exp(3<em>x</em>)

= 2<em>x</em> + (1 + 2<em>x</em>) exp(<em>x</em>) + 4 exp(3<em>x</em>)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

<em>x</em> : 2<em>a</em> = 2

1 : -3<em>a</em> + 2<em>b</em> = 0

exp(<em>x</em>) : 2<em>c</em> - <em>d</em> = 1

<em>x</em> exp(<em>x</em>) : -2<em>c</em> = 2

exp(3<em>x</em>) : 2<em>e</em> = 4

Solving the system gives

<em>a</em> = 1, <em>b</em> = 3/2, <em>c</em> = -1, <em>d</em> = -3, <em>e</em> = 2

Then the general solution to the DE is

<em>y(x)</em> = <em>C₁</em> exp(2<em>x</em>) + <em>C₂</em> exp(<em>x</em>) + <em>x</em> + 3/2 - (<em>x</em> ² + 3<em>x</em>) exp(<em>x</em>) + 2 exp(3<em>x</em>)

4 0
3 years ago
POSSIBLE POINTS: 5.88
Bess [88]

The function of the length z in meters of the side parallel to the wall is A(z) = z/2(210 - z)

<h3>How to write a function of the length z in meters of the side parallel to the wall?</h3>

The given parameters are:

Perimeter = 210 meters

Let the length parallel to the wall be represented as z and the width be x

So, the perimeter of the fence is

P = 2x + z

This gives

210 = 2x + z

Make x the subject

x = 1/2(210 - z)

The area of the wall is calculated as

A = xz

So, we have

A = 1/2(210 - z) * z

This gives

A = z/2(210 - z)

Rewrite as

A(z) = z/2(210 - z)

Hence, the function of the length z in meters of the side parallel to the wall is A(z) = z/2(210 - z)

Read more about functions at

brainly.com/question/1415456

#SPJ1

5 0
1 year ago
What is the equation of a line that passes through the point(3,-8)and has a slope of 0?
hram777 [196]

Answer: y=-8

Step-by-step explanation:

If the slope is zero that means that the line does not go up (or have any rise) and therefore must be a horizontal line. Since it passes through the point (3,-8), the equation has to be y=-8.

7 0
2 years ago
What precent of 25 is 10?​
vivado [14]

40%

40 percent of 25 is 10

10/25x100 and you will get your answer which is 40

4 0
2 years ago
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