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3 years ago

I'm so sorry for whoever answers this, I am really having a brain fart plz help

Mathematics

2 answers:

Nikolay [14]3 years ago

7 0

Answer:

Step-by-step explanation:

it would be 1,999,700

MORE can i have brainlyest

agasfer [191]3 years ago

4 0

Answer: 1,999,000

Step-by-step explanation:

And yes I made sure it was right

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klio [65]

Answer:

Step-by-step explanation:

You want to factor out perfect squares if they exist.

$1.\ \sqrt{36}=\sqrt{6^2}=6\\ \\ 2.\ \sqrt{40}=\sqrt{2^2(10)}=2\sqrt{10}\\ \\ 3.\ \sqrt{75}=\sqrt{3(5^2)}=5\sqrt{3}\\ \\ 4.\ \sqrt{120}=\sqrt{2^2(30)}=2\sqrt{30}\\ \\ 5.\ \sqrt{84}=\sqrt{2^2(21)}=2\sqrt{21}\\ \\ 6.\ \sqrt{48}=\sqrt{4^2(3)}=4\sqrt{3}$

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3 years ago

Read 2 more answers

. You draw a card from a well-shuffled deck of 52 cards. If you get a red card, you win nothing. If you get a spade, you win $4.

Novosadov [1.4K]

Answer is in the file below

tinyurl.com/wtjfavyw

7 0

3 years ago

Help pls I'll give 25 points

RideAnS [48]

Answer:

$\text{C. }60$

Step-by-step explanation:

<u>Question from image</u>:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the <em>x</em> marks where it cannot be.

$\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}$

Let's start with the first position:

$\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}$

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in $3!=6$ ways. Therefore, there are $4\cdot 6=\boxed{24}$ possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

$\underline{X}\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}$

There are three places the 2 can be, since the 2 must be behind the 1. For each of these three places, the remaining 3 digits can be arranged in $3!=6$ ways. Therefore, there are $3\cdot 6=\boxed{18}$ possible numbers when 1 is the second digit of the number.

The pattern continues. Next there will be 2 places to place the 2. For each of these, there are $3!=6$ ways to rearrange the remaining 3 digits for a total of $2\cdot 6=\boxed{12}$ possible numbers when 1 is the third digit of the number.

Lastly, when 1 is the fourth digit of the number, there is only 1 place the 2 can be. For this one place, there are still $3!=6$ ways to rearrange the remaining three numbers. Therefore, there are $1\cdot 6=\boxed{6}$ possible numbers when 1 is the fourth digit of the number.

Thus, there are $24+18+12+6=\boxed{60}$ numbers that will have the digits 1 and 2 in increasing order, from a set of 120 five-digit numbers created by the digits 1 through 5, where no digit may be repeated.