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mixas84 [53]
4 years ago
11

If John makes $20 an hour, how long will it take him to make $140?

Mathematics
1 answer:
Len [333]4 years ago
4 0
For 1 hour, money = $20
For $140, it would be: 140/20 = 7 hours.

So, option C is your final answer.

Hope this helps!
You might be interested in
M is directly proportional to r2.when r=2,m=14.Work out the value of m when r=12
valentinak56 [21]

Answer:

504

Step-by-step explanation:

In the attached file

Hope it helps

4 0
4 years ago
Write the sum as a product simplify the product-2 + -2 + -2 + -2 equals
yuradex [85]
-2+-2+-2+-2
-4+-4
=-8
hope this helps:)
6 0
3 years ago
Kyle mows4 lawns a day 4 day a week 4 week a month and 4 months a year he makes $25 per lawn how much money did he make last yea
denis-greek [22]

Lets break this down step by step.

4+4+4+4=16 lawns a week.

16+16+16+16=64 lawns a month.

64+64+64+64=256 lawns a year.

Then we take 256 x $25 to get $6,400 last year.

6,400<---Answer

-Seth

8 0
3 years ago
What is the 50th term of the arithmetic sequence 3, 7, 11, 15, ... ?
zysi [14]

Answer:

B. 199

Step-by-step explanation:

We are given an Arithmetic sequence : 3, 7, 11, 15, ...

We are supposed to find the 50th term .

So, first consider the formula of nth term in Arithmetic mean:

a_{n} = a+(n-1)*d

where

a is first term of sequence

n is term position

d is common difference

a_{n}= the term you are supposed to find

Now,

a = 3 ( first term )

n = 50 th term ( given )

d = 7-3 = 11-7 = 4(common difference

Substitute these value in the formula given above

⇒a_{50} = 3+(50-1)*4

⇒a_{50} = 3+(49)*4

⇒a_{50} = 3+196

⇒a_{50} = 199

Hence ,The 50th term of given arithmetic sequence is 199

So, OPTION B is correct .


5 0
3 years ago
PLEASE HELP ON #3 ASAP!
IrinaK [193]

9514 1404 393

Answer:

  1a: x+3 = 5

  1c: 6 = 2z

  2b: x = 2

  2d: 3 = z

  3: the solutions make the hangars balance

Step-by-step explanation:

1. We can write the equations by listing the contents of the hangar and using an equal sign to show the balance between left side and right side. It can work well to put left side contents of the hangar on the left side of the equal sign.

  A: x + 3 = 5

  C: 1 + 1 + 1 + 1 + 1 + 1 = z + z  simplifies to  6 = 2z

__

2. B: We can subtract 3 from both sides of the hangar (and equation) to find the value of x.

  (x +3) -3 = 5 -3

  x = 2 . . . . . hangar balances with 2 on the right

D: We can divide both sides of the hangar by 2, splitting the content into two equal parts. Then one of those parts can be removed from each side.

  2(3) = 2(z)

  3 = z . . . . . . hangar balances with 3 on the left

__

3. The found values will keep the hangar in balance when they are substituted for the corresponding variables.

  A: 2 + 3 = 5

  C: 1 + 1 + 1 + 1 + 1 + 1 = 3 + 3

4 0
3 years ago
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