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Vesnalui [34]
1 year ago
8

What is the first step in subtracting fractions??

Mathematics
2 answers:
Elden [556K]1 year ago
8 0

Answer:

make sure they have the same denominator

Ipatiy [6.2K]1 year ago
4 0
Find a common denominator
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The price of a new car was 12500 it is reduced 11625 work out the percentage reduction
mart [117]
% Decrease = Decrease ÷ Original Number × 100

Divde 11,625 by 12,500 and multiply the answer by 100.

11,625 ÷ 12,500 × 100 = 93 

The percentage reduction is 93%.
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3 years ago
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Alice sold candy for a school fundraiser. Chocolate bars sold for$2.25 a piece, bags of chocolate covered peanuts sold for$1.85
Rzqust [24]

Answer:

12 bags

Step-by-step explanation:

2.25 × 9 = 20.25

.50 × 10 = 5

20.25 + 5 = 25.25

47.45 - 25.25= 22.2

22.2 ÷ 1.85 = 12

8 0
3 years ago
PLEASE HELP FOR BRAINLIEST ANSWER I NEED TO MAKE SURE THIS IS CORRECT
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My other answer got deleted.. I did answer your question, but whatever.

You are correct, the answer is D. a reflection across y = x then a positive rotation of 270 degrees about the origin.
8 0
4 years ago
Write three different sentences that could describe the relationship between the quantities $27, $81
Olenka [21]
1) 27+x=81
2)27+81=108
3)81-x=27
8 0
4 years ago
In the expansion of ( x^3 - 2/x^2 ) ^10 , find the coefficient of 1/x^5​
Lilit [14]

Answer:

240

Step-by-step explanation:

We need to find the coeffeicent of the binomial expansion of

( {x}^{3}  - 2 {x}^{ - 2} ) {}^{10}

Note that

- 2 {x}^{ - 2}  = -  \frac{2}{ {x}^{2} }

The binomial theorem states that

(x + y) {}^{n}  = x {}^{n} y {}^{0}  +  \binom{n}{1} x {}^{n - 1} y +  \binom{n}{2} x {}^{n - 2} y {}^{2} ....... + x {}^{0} y {}^{n} ( \binom{n}{n} )

Using this, we let expand our series

( {x}^{3}   - 2 {x}^{ - 2} ) {}^{10}  = x {}^{30}  +  \binom{10}{1} ( {x}^{27}     2 {x}^{ - 2} ) +  \binom{10}{2}  {x}^{24} 2x {}^{ - 4}  +  \binom{10}{3}  {x}^{21} 2x {}^{ - 6}  +  \binom{10}{4}  {x}^{18} 2x { }^{ - 8}  +  \binom{10}{5} x {}^{15} 2x {}^{ - 10}  +  \binom{10}{6} x {}^{12}2 x {}^{ - 12}  +  \binom{10}{7} x {}^{ 9} 2x {}^{ - 14}  +  \binom{10}{8} x {}^{ 6} 2x {}^{ - 16}  +  \binom{10}{9} ( {x}^{3} )2x {}^{ - 18}  + 2x {}^{ - 20}

\frac{1}{ {x}^{5} }  = x {}^{ - 5}

So what term in the series eqaul x^-5.

That term is the 10 choose 7 term.

\binom{10}{7}  {x}^{9} 2x {}^{ - 14}

Because

=  \binom{10}{7} 2x {}^{ - 14}  {x}^{9}  =  \binom{10}{7} 2 {x}^{ - 5}

So we need to compute 10 choose 7.

That equals

10!/3!(7!)= 10×9×8/6= 720/6=120.

So we get

120(2) {x}^{ - 5}

240 {x}^{ - 5}

So the coeffceint u

is 240

3 0
3 years ago
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