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3 years ago

A pizza shop charged $49.57 for 3 large pepperoni pizzas. This price included a $2.50 delivery fee.

Mathematics

2 answers:

Amiraneli [1.4K]3 years ago

5 0

It will be $49.57 - $2.50 = $47.07

tia_tia [17]3 years ago

4 0

Answer:

$15.69

Step-by-step explanation:

Lets make an equation...

Let p = the price of one pizza.

$49.57 Divided by 3 = ?.

? = 16.523333333333_ Repeating.

Simplify it to 16.52.

16.52 is not an answer choice.

Solution:

Subtract the 2.50 by 49.57 to get 47.09.

Recreate the equation.

47.09 Divided by 3 = ?.

? = 15.693333333333_ Repeating.

Simplify it to 15.69.

p = 15.69.

Hope i helped!

-Snooky73hh

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Use the equation and type the ordered-pairs. y = log 3 x {(1/3, a0), (1, a1), (3, a2), (9, a3), (27, a4), (81, a5)

vagabundo [1.1K]

Answer:

Considering the given equation $y = log_{3}x\\$

And the ordered pairs in the format $(x, y)$

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For $(\frac{1}{3}, a_{0} )$

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$y = log_{3}x\\y = log_{3}1\\y = log_{3}(1)\\Note: \log _a(1)=0\\y = 0$

So the ordered pair will be $(1, 0 )$

For $(3, a_{2} )$

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$y = log_{3}x\\y = log_{3}3\\y = 1$

So the ordered pair will be $(3, 1 )$

For $(9, a_{3} )$

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So the ordered pair will be $(9, 2 )$

For $(27, a_{4} )$

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$y = log_{3}x\\y = log_{3}27\\y=3\log _3\left(3\right)\\y=3$

So the ordered pair will be $(27, 3 )$

For $(81, a_{5} )$

$x=81$

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4 years ago

Jeffrey went to the movies with his friends. The tickets cost $8.50 each for students and they spent $15.00 altogether at the co

LenaWriter [7]

Answer:

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Step-by-step explanation:

Let's first write an equation.

We want to find the number of tickets t

The total was $49 so our equation will equal $49 and they spent $15 in snacks so on the other side of the eqaution we have to add $15, on the same side as the snacks we need to add in the cost of tickets, which is $8.50 multiplies by t, or the total number of tickets.

$49 = $8.50t + $15

Solve for t, first subtract $15 from both sides.

$49 - $15 = $8.50t

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$34/$8.50 = $8.50t/$8.50

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4 tickets meaning 4 people went.

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3 years ago

Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR su

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<h2>Answer with explanation:</h2>

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f: R → R is a continuous function such that:

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i.e.

f(0)=f(0)+f(0)

By using property (1)

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f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0 )

Also,

f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1) ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

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f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

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f(m)=mk

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Also,

when x∈ Q

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