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12345 [234]
3 years ago
12

This ones hard for me

Mathematics
2 answers:
nataly862011 [7]3 years ago
8 0
The answer is no most likely
Nonamiya [84]3 years ago
3 0
The answer is no , because there is 0 twice !
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The underlined digit is 3 wat the value of this number <br> 285,003
Triss [41]

Answer:

just 3

Step-by-step explanation:

it would be just 3 because there is no number behind it

4 0
3 years ago
Amino Buys A Book For Rupees 275 And Sells It At A Loss Of 15% How Much Does She Sell It For
Kruka [31]

Answer:

$233.75

Step-by-step explanation:

2.75 x 15 = 41.25

275 - 41.25 = $233.75

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3 years ago
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What is a synonym for shrek’s penis?
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Solve the equation using the zero-product property. (2x − 8)(7x + 5) = 0 x = –2 or x = 7 x = 4 or x = x = 4 or x = x = –4 or x =
aliya0001 [1]

(2x-8)(7x+5)=0\iff2x-8=0\ \vee\ 7x+5=0\\\\2x-8=0\qquad\text{add 8 to both sides}\\2x=8\qquad\text{divide both sides by 2}\\\boxed{x=4}\\\\7x+5=0\qquad\text{subtract 5 from both sides}\\7x=-5\qquad\text{divide both sides by 7}\\\boxed{x=-\dfrac{5}{7}}

5 0
3 years ago
Amy's grandmother gave her 3 identical chocolate chip cookies and 4 identical sugar cookies. In how many different orders can Am
Dmitrij [34]

Answer:

chocolate chip cookie first: 15 ways

chocolate chip cookie last: 15 ways

chocolate chip cookie first and last: 5 ways

Step-by-step explanation:

There is a total of 7 cookies.

Is she eats a chocolate cookie first, she will have 2 chocolate cookies and 4 sugar cookies (6 in total).

So, to find the number of different orders that Amy can eat the remaining cookies, we just need to calculate a combination of 6 choose 2 (that is, the number of ways we can put the 2 chocolate cookies among the 6 cookies) or a combination of 6 choose 4 (same thing, but for the sugar cookies):

C(6,2) = 6! / (2! * 4!) = 6 * 5 / 2 = 15 ways

Is she eats a chocolate cookie last, she will have 2 chocolate cookies and 4 sugar cookies (6 in total), so the problem is solved again with a combination of 6 choose 2:

C(6,2) = 15 ways

Is she eats a chocolate cookie first and last, she will have just 1 chocolate cookie left and 4 sugar cookies (5 in total), so the problem is solved with a combination of 5 choose 1 or a combination of 5 choose 4:

C(5,1) = 5! / (1! * 4!) = 5 ways

6 0
3 years ago
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