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3 years ago

Where do I put the parentheses in 21-8-6=7

Mathematics

2 answers:

Sliva [168]3 years ago

7 0

Answer:

around 21-8 so the equation would look like this (21-8)-6=7

Step-by-step explanation:

you do this because if u put the parentheses around 8-6 you would get 21- (8-6)= 19 because 21 - 2 = 19 so you have to put them around 21-8 to get the right answer

Pavel [41]3 years ago

4 0

Answer:

(21-8)-6=7

Step-by-step explanation:

If you put the parentheses holding the 8-6 it makes the statement equal 19 and the other way is true so ya hope this helps <3

Also can you give brainliest to all of your questions bc it really hard to get brainliest and it be really nice you can choose who i dont really care but just please do it it helps a lot

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8.45 At some point in the season, the Unicorns had won 60% of their basketball games. After that point, they won 8 more games an

ololo11 [35]

Answer:

The total number of games won by Unicorn is 40.

Step-by-step explanation:

We are given with a word problem
We are asked to find the number of unicorns played during the season
We can do this in two steps

Step 1: Finding the win percentage

Step 2: Finding the number of unicorns played during the season

Step 1 of 2

Let the number of games played by unicorn be x.The unicorn won 60% of the first x-10 games That is $\frac{60}{100} \times(x-10)$

Combining the 8 games we get,

$\begin{gathered}\frac{60}{100}(x-10)+8=\frac{60}{100} \times x-\frac{60}{100} \times 10+8 \\0.6 x-6+8=0.6 x+2\end{gathered}$

Step 2 of 2

The unicorn totally won 65% of the games played.

That is $\frac{65}{100} \times x=0.65 x$

Equating both the equation gives

$\begin{gathered}0.6 x+2=0.65 x \\2=0.65 x-0.6 x \\2=0.05 x \\\frac{2}{0.05}=x \\40=x\end{gathered}$

5 0

2 years ago

Six friends attend a party. They form pairs for a game. How many different pairs are possible?

Leona [35]

Pairs, in this case, relates to a group of 2 or more. We have 6 friends. Let's call them A,B,C,D,E,F. This will allow us to make a [some sort of] combination tree:

1. ABC against DEF

2. ABD against CEF

3. ABE against CDF

4. ABF against CDE

5. ACD against BFE

6. ACE against BDF

7. ACF against BDE

8. ADE against BCF

9. ADF against BCE

10. AEF against BCD

I believe there are 12 combinations... I just can't think of the last 2 though.

5 0