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Anni [7]
3 years ago
6

A taxi service offers a ride with an $5 sub charge and charges 0.50 per mile

Mathematics
1 answer:
Lynna [10]3 years ago
4 0

Answer:

5+.5(M)

Step-by-step explanation:

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In a different plan for area​ codes, the first digit could be any number from 2 through 5​, the second digit was either 2, 3, 4,
GrogVix [38]

Answer:

Step-by-step explanation:

First digit 2,3,4,5 = 4 different digits.

Second digit 2,3,4,5 = 4 different digits.

Third digit = 1,2,3,6,7,9 = 6 different digits.

Total number of possibilities = 4*4*6 = 96

3 0
3 years ago
Using 50 boxes of nails a carpenter a
lapo4ka [179]
450=50x this should be correct :)
8 0
3 years ago
WILL GIVE BRAINLIEST The sequence 7 21 63 189......shows the number of jumping jacks justin did each week, starting with the fir
vovangra [49]

A

this is a geometric sequence since there exists a common ratio r between the terms

r = \frac{21}{7} = \frac{63}{21} = \frac{189}{63} = 3

B

to obtain the next term in the sequence multiply the previous term by 3

a_{n+1} = 3 a_{n} ← recursive rule

C

the n th term of a geometric sequence is

a_{n} = a_{1} r^{n-1}

where a_{1} is the first term in the sequence

a_{n} = 7 × 3^{n-1} ← explicit rule


3 0
3 years ago
FAST ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
lana [24]

Answer:

0.5

Step-by-step explanation:

just use any point you can find in the graph

and use the dollar to divided by the cups.

the point I chose is 1 dollar for 2 cups. So it's 0.5 dollar for 1cup

5 0
3 years ago
X4y(4) = yx2(13), calculate (x+y)2
Delicious77 [7]

Please, for clarity, use " ^ " to denote exponentiation:


Correct format: x^4*y*(4) = y*x^2*(13)

This is an educated guess regarding what you meant to share. Please err on the side of using more parentheses ( ) to show which math operations are to be done first.


Your (x+y)2, better written as (x+y)^2, equals x^2 + 2xy + y^2, when expanded.

The question here is whether you can find this x^2 + 2xy + y^2 in your

"X4y(4) = yx2(13)"


Please lend a hand here. If at all possible obtain an image of the original version of this problem and share it. That's the only way to ensure that your helpers won't have to guess what the problem actually looks like.


3 0
4 years ago
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