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Nutka1998 [239]
4 years ago
10

Math i have $92 how much i need to get to $150

Mathematics
2 answers:
Viefleur [7K]4 years ago
6 0
68 is how much more money you need

Alisiya [41]4 years ago
5 0
150$ - 92$ = 58$

This is how you can find your awnserto your question.
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Apply the distributive property to factor out the greatest common factor.<br> 55+35
pochemuha
To factor out the greatest common factor, you need to first list them out:
35: 1, 5, 7, 25
55: 1, 5, 11, 55

As you can see, the greatest common factor here is 5.
Which means:
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3 years ago
Represent the following expressions as a power of the number a (a +0)<br> a^2 * (a^0)^3/a^11
LekaFEV [45]
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3 years ago
How many integers between 2020 and 2400 have four distinct digits arranged in increasing order?
malfutka [58]

Answer:  E=15

There are 15 integers between 2020 and 2400 which have four distinct digits arranged in increasing order.

Step-by-step explanation:

This can be obtained by after a simple counting of number from 2020 and 2400 as follows:

The first set of integers are:

2345, 2346, 2347, 2348, and 2349.

Therefore, there are 6 integers in first set.

The second set of integers are:

2356, 2357, 2358, and 2359.

Therefore, there are 4 integers in second set.

The third set of integers are:

2367, 2368, and 2369.

Therefore, there are 3 integers in third set.

The fourth set of integers are:

2378, and 2379.

Therefore, there are 2 integers in fourth set.

The fifth and the last set of integer is:

2389

Therefore, there is only 1 integers in fifth set.

Adding all the integers from each of the set above, we have:

Total number of integers = 6 + 4 + 3 + 2 + 1 = 15

Therefore, there are 15 integers between 2020 and 2400 which have four distinct digits arranged in increasing order.

Sorry if it is incorrect

8 0
3 years ago
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Answer:

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