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Eddi Din [679]
3 years ago
11

If your income is ​$27 a​ month, the price of pizza is​ $5 and the price of a video is​ $4, how many pizzas will you buy and how

many videos will you rent each​ month?
a. 3 pizzas and 3 videos.
b. 4 of each.
c. 5 pizzas and 4 videos.
d. 6 pizzas and 2.5 videos.
Mathematics
2 answers:
n200080 [17]3 years ago
7 0
Hello!

For this you can do multiplication for each of the answers

for A: 5*3 = 15 and 4*3 = 12   15 + 12 = 27

B: 5*4 = 20 and 4*4 = 16   20+16 = 36

The answer is A

Hope this helps!
ELEN [110]3 years ago
7 0
D is the answer for this question
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Hope this helps please mark this as brainliest answer
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Holly begins with 1 penny in her piggy bank. She adds 2 pennies on day 1, 4 pennies on day 2, and 8 pennies on day 3. She contin
muminat

Answer:

All the options are correct.

Step-by-step explanation:

Holly begins with 1 penny in her piggy bank. She adds 2 pennies on day 1, 4 pennies on day 2, and 8 pennies on day 3. She continues this pattern for 30 days.

The pattern which she choses is 2^n where n denotes the number of days.

A)

The graph of this sequence would pass the vertical line test.

" Vertical line test: A function can only have one output, y, for each unique input, x " .

Here also y=2^n and x=n.

B)

Each day corresponds to a unique amount of pennies in this sequence.

Since the sequence is increasing, hence option B is correct.

C)

The graph of this sequence would pass the horizontal line test.

A test use to determine if a function is one-to-one.

As for each n we get a unique value.

Hence, option C is correct.

D)

Each amount of pennies corresponds to a unique day in this sequence.

This option is also true as each  amount is a function of number of days.

5 0
3 years ago
Read 2 more answers
Find the distance between<br> 2-3i<br> And<br> 9+21i
Goryan [66]

2 - 3i is just another of writing (2, -3) in the cartesian plane, the 2-3i however is using the "imaginary axis" for the imaginary plane, is all however the imaginary axis is simply the equivalent of the y-axis.  Likewise 9+21i is just (9,21), so let's just use the distance formula.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{9}~,~\stackrel{y_2}{21})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[9-2]^2+[21-(-3)]^2}\implies d=\sqrt{(9-2)^2+(21+3)^2} \\\\\\ d=\sqrt{7^2+24^2}\implies d=\sqrt{49+576}\implies d=\sqrt{625}\implies d=25

3 0
3 years ago
Find the indefinite integral. (Note: Solve by the simplest method—not all require integration by parts. Use C for the constant o
umka2103 [35]

Answer:

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2  ln⁡(1+x^2 )+C

Step-by-step explanation:

∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗

Let 1st=arctan⁡(x)

And 2nd=1

∫▒〖arctan⁡(x).1 dx=arctan⁡(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗

As we know that  

derivative of arctan(x)=1/(1+x^2 )

∫▒〖1 dx〗=x

So  

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1

Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now  

Let 1+x^2=u

du=2xdx

Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get

1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)  

1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)  

1/2 ∫▒(2xdx/u) =1/2  ln⁡(u)+C

1/2 ∫▒(2xdx/u) =1/2  ln⁡(1+x^2 )+C

Putting values in Eq1 we get

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2  ln⁡(1+x^2 )+C  (required soultion)

3 0
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Please answer this correctly
tekilochka [14]
Take the coordinates
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