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saul85 [17]
3 years ago
9

Pls help ill give brainliest

Mathematics
1 answer:
Bond [772]3 years ago
3 0
It is located in quadrant 2, as it is originally in the third quadrant and is flipped over the x axis.
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How does the 2 in the hundredths place compare to the 2 in the thousandths place?
dybincka [34]

Answer:

2 in the hundredths place is greater

Step-by-step explanation:

2 in the hundredths place is 10 times the thousands place

7 0
4 years ago
The coordinates of the vertices of the triangle shown below are a(2,13), b(10,5) c(12,14).
Luden [163]
(2,13)is the right answer
5 0
3 years ago
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Which graph would you want to use if you worked for Brand A and wanted to show that your product is preferred almost as often as
jasenka [17]
It would be C. The right one makes it look horrible compared to the other, if someone did not look at the graph.
5 0
4 years ago
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41=12d-7 can someone help me?​
OverLord2011 [107]

Answer:

d = 4

Step-by-step explanation:

41 = 12d - 7

48 = 12d

d = 4

6 0
4 years ago
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Ayesha has a theory about two digit numbers that have a larger tens digit than ones digit. she says that reversing the digits an
stira [4]
<span>Ayesha's right.  There's a good trick for knowing if a number is a multiple of nine called "casting out nines."  We just add up the digits, then add up the digits of the sum, and so on.  If the result is nine the original number is a multiple of nine.  We can stop early if we recognize if a number along the way is or isn't a multiple of nine.  The same trick works with multiples of three; we have one if we end with 3, 6 or 9.

So </span>1284673 <span>has a sum of digits 31 whose sum of digits is 4, so this isn't a multiple of nine.  It will give a remainder of 4 when divided by 9; let's check.

</span>1284673 = 142741 \times 9 + 4 \quad\checkmark<span>

</span>Let's focus on remainders when we divide by nine. The digit summing works because 1 and 10 have the same remainder when divided by nine, namely 1.  So we see multiplying by 10 doesn't change the remainder.  So 100 \times a + 10 \times b + c has the same remainder as a+b+c.

When Ayesha reverses the digits she doesn't change the sum of the digits, so she doesn't change the remainder.  Since the two numbers have the same remainder, when we subtract them we'll get a number whose remainder is the difference, namely zero. That's why her method works.
<span>
It doesn't matter if the digits are larger or smaller or how many there are. We might want the first number bigger than the second so we get a positive difference, but even that doesn't matter; a negative difference will still be a multiple of nine. Let's pick a random number, reverse its digits, subtract, and check it's a multiple of nine:

</span>813219543092195 - 591290345912318 =  221929197179877 = 9 \times 24658799686653 \quad \checkmark

6 0
3 years ago
Read 2 more answers
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