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Veronika [31]
3 years ago
9

Can someone help me with these math problems please? I need it done by tonight if someone can please help me by then...I'll make

the one with the right answers brainliest. Don't answer just for the points.. So if anyone can please help me it'll be helpful please and thanks must have right answers and done by tonight..

Mathematics
1 answer:
kotegsom [21]3 years ago
4 0

Answer:

For question 3, you would just add 2 to the x values and subtract 2 from the y values, so it would be:

J' (-2, 5)

K' (2, 6)

L' (1, 2)

M' (-3, 1)

For question 4 you would subtract 7 from the x values and 6 from the y values, and that would be:

W' (-6, 1)

X' (-1, -1)

Y' (-3, -6)

Z' (-8, -4)

For question 9 you would end up with:

X' (6, -5)

Y' (7, 1)

Z' (4, 0)

For question 10 you would end up with:

Q' (-1, 2)

R' (1, 7)

S' (-2, 6)

T' (-4, 1)

For question 11 you would end up with:

L' (4, 1)

M' (8, 5)

N' (6, 7)

P' (2, 3)

For question 12 you would end up with:

G' (6, -7)

H' (6, -4)

I' (1, -7)

Hope this is what you were looking for!

 

Step-by-step explanation:


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How many three digit numbers have digits whose sum is greater than 2?
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Answer:

896

Step-by-step explanation:

Let's talk first about how many 3 digit numbers there are. The first 3 digit number is 100 and the last is 999. So there are 999-100+1 numbers that are 3 digits long. That simplifies to 900.

Now let's find how many of those have a sum for the digits being 1, then 2 ? Then take that sum away from the 900 to see how many 3 digit numbers have the sum of their digits being more than 2.

3 digit numbers with sum of 1:

The first and only number is 100 since 1+0+0=1.

We can't include 010 or 001 because these aren't really three digits long.

3 digit numbers with sum of 2:

The first number is 101 since 1+0+1=2.

The second number is 110 since 1+1+0=2.

The third number is 200 since 2+0+0=2.

That's the last of those. We could only use 0,1, and 2 here.... Anything with a 3 in it would give us something larger than or equal to 3.

So there are 900-1-3 numbers who are 3 digits long and whose sum of digits is greater than 2.

This answer simplifies to 896.

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