2 consecutive odd integers whose product is 63

Mathematics

2 answers:

disa [49]3 years ago

8 0

Check: 7*9=63

There you go!

Galina-37 [17]3 years ago

3 0

To solve it you would make an equation where x*(x+2)=63 where x is the smaller integer. this is because an odd number is always 2 away from the odd number after it. Multiply it out to get x^2+2x=63 and subtract 63 from both sides to get a trinomial. X^2+2x-63 factors into (x+9)(x-7) this means your smaller integer is either -9 or 7 making your larger integers being -7 and 9. Your two pairs of integers are: (7 and 9) and (-9 and -7) :)

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Place values for 249.673

irakobra [83]

Hi!

This question is worded a little funny, but i think in understand what it's asking. Forgive me if i'm wrong.

Place values in math are pretty self-explanatory. It's just asking you what place each number is in.

For example:

If you were to write the number 12, the 1 would be in the tens place and the 2 would be in the ones place.

When speaking about decimals, you pretty much just have to add the letters 'th' to the end of the place value, but instead of decreasing as the numbers grow smaller, you increase and the ones place doesn't exist.

So in this case:

With 249.637, 2 is in the hundreds place. 4 is in the tens place. 9 is in the ones place. 6 is in the tenths place. 7 is in the hundreths place. 3 is in the thousandths place.

Hope this helps :)

3 0

3 years ago

Each number below is a factor of Miguel's number.

BabaBlast [244]

Pretty sure it d.12 bc you were counting from 3’s so the next number up would be 12

6 0

3 years ago

Read 2 more answers

HELP PLSSS THIS IS HARD SOMEONE

tia_tia [17]

Answer:

x,y - x,y-4