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IrinaVladis [17]
3 years ago
10

Answer this for me please

Mathematics
2 answers:
Arada [10]3 years ago
8 0

Answer: C

Step-by-step explanation:

C becuase they kept the fee so thats +4 so now were looking for x. Last year it was 6 and they doubled it so its 12

Leona [35]3 years ago
6 0

Answer:

C, y = 12x + 4

Step-by-step explanation:

If they double the ticket price, aka 6, then it would be 12. They also kept the convenience charge the same, so that remains 4.

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20 points! Clear answers and good descriptions!<br> Thanks! :)
balu736 [363]

Answer:

1= b

2= f

Step-by-step explanation:

8 0
2 years ago
The Oregon trail is 2.197 miles long. How long would it take a covered wagon travelling 20 miles a day to complete the trip? Wri
Bess [88]

Answer: It needs 2.4 hours to complete the trip.

Step-by-step explanation:

Since we have given that

Length of Oregon trail = 2.197 miles

Speed at which it travel = 20 miles per day

we need to find the time taken by it ,

As we know that

Time=\frac{Distance}{Speed}

Time=\frac{2.197}{20}\\\\Time=0.10\text{ days}

So,

it will be changed into hours

1\text{ day}=24\text{ hours}\\\\0.1\text{ day}=24\times 0.1=2.4\text{ hours}

so, it needs 2.4 hours to complete the trip.


3 0
3 years ago
Why do I have to go to school after the 16 I thought that’s when we don’t goats?
Y_Kistochka [10]
Because school is lame
6 0
3 years ago
450 is 75% of what number
Sergeu [11.5K]
•<span>75% = 0.75
•</span><span>450 ÷ 0.75 = 600
•</span><span>450 is 75% 0f 600
There are the steps that I used to get that </span><span>450 is 75% 0f 600</span>
3 0
3 years ago
Read 2 more answers
Petes boat can travel 48 miles upstream in 4 hours. The return trip takes 3 hours. Find the speed of the boat in still water and
tankabanditka [31]
So... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up

going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r

notice, the distance is the same, upstream as well as downstream
thus   \bf \begin{cases}&#10;b=\textit{rate of the boat}\\&#10;r=\textit{rate of the river}&#10;\end{cases}\qquad thus&#10;\\\\\\&#10;&#10;\begin{array}{lccclll}&#10;&distance&rate&time(hrs)\\&#10;&----&----&----\\&#10;upstream&48&b-r&4\\&#10;downstream&48&b+4&3&#10;\end{array}&#10;\\\\\\&#10;&#10;\begin{cases}&#10;48=(b-r)(4)\to 48=4b-4r\\\\&#10;\frac{48-4b}{-4}=r\\&#10;--------------\\&#10;48=(b+r)(3)\\&#10;-----------------------------\\\\&#10;thus\\\\&#10;48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3)&#10;\end{cases}

solve for "r", to see what the stream's rate is

what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
5 0
3 years ago
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