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FrozenT [24]
3 years ago
12

Chris made $184 for 8 hours of work. At the same rate, how much would he make for 13 hours of work?

Mathematics
2 answers:
masha68 [24]3 years ago
6 0

Answer:

Step-by-step explanation: 184 ÷8 =23 per hour

So then if 23x13=299

pashok25 [27]3 years ago
4 0

Answer:

$299

Step-by-step explanation:

Make a proportion

He makes $184 for 8 hours of work, and $x for 13 hours of work

184/8=x/13

To solve, we need to get x by itself. To do this, multiply both sides by 13. We multiply because x is being divide by 13, and we want to do the opposite.

13(184/8)=(x/13)13

The 13s on the right will cancel, leaving us with

13(23)=x

299=x

So, he will earn $299 for 13 hours of work.

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The vertices of a square are located at:
vivado [14]

Answer:

(4,4), (10,4), (10,10), (4,10)

Step-by-step explanation:

(2,2) x2 -> (4,4)

(5,2) x2 -> (10,4)

(5,5) x2 -> (10,10)

(2,5) x2 ->  (4,10)

Hope that helps!

5 0
3 years ago
Read 2 more answers
One more than three times a number is less than the number decreased by five. solve and graph
Genrish500 [490]
Whats the graph ??
 i need it to find the answer


4 0
3 years ago
−1/2 (x+2) + 112x = 3<br><br> x=?
AysviL [449]

Given equation :−1/2 (x+2) + 112x = 3

We have -1/2 in front of parenthesis on left side .

It's better to remove fraction in an equation, in order make it easier to solve.

In order to remove 2 from denominator of 2, we need to multiply each and every term by 2.

Multiplying each term in the equation by 2, we get

2* −1/2(x+2) +2* 112x = 2*3.

On simplifying this step, we get

-1(x+2) +224x = 6.

Distributing -1 over (x+2), we get

-x -2 +224x = 6

Combining like terms on left side, -x+224x=223x

223x -2 = 6

Adding 2 on both sides, we get

223x -2+2 = 6+2

223x = 8

Dividing both sides by 223, we get

223x/223 = 8/223.

x= 8/223.

6 0
3 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

4 0
3 years ago
What does 1/10 of 3000 equal?
arsen [322]
Answer:-
1/10×3000
=300
4 0
3 years ago
Read 2 more answers
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