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sasho [114]
3 years ago
15

Y = 4X-3 4.88-1:2y = 3.6

Mathematics
1 answer:
Rzqust [24]3 years ago
7 0

Answer:

google it , it'll help you

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bonufazy [111]
The unsimplified answer is 15/50. The answer simplified is 3/10.
equation/how to solve=
there are 50 circus performers total. (15 magicians + 35 performers) so that would be 15 magiciand over 50 performers (15/50)
to simplify... 5 goes into 15 3 times, and 5 goes into 50 ten times. (3/10)
Hope this helped you out!
6 0
3 years ago
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If a=3/6 and b=-3/-5 find -3a+4b
Fiesta28 [93]

Answer:

Step-by-step explanation:

-3(3/6) + 4(3/5)

-3(1/2) + 4(3/5)

-3/2 + 12/5

-15/10 + 24/10

9/10

7 0
3 years ago
Can somebody help me please
Sergeu [11.5K]
Area = length x width

24 = (n-3) x 2
24 = 2n -6
2n = 30
n = 15

Length is 15-3 =12, width is 2
24 = 12 x 2
3 0
3 years ago
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Please help thanks! Brainliest
KatRina [158]
9 x 9 = 81
81 - 21 = 60
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4 0
2 years ago
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What is the expansion of (3+x)^4
Vlad1618 [11]

Answer:

\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81

Step-by-step explanation:

Considering the expression

\left(3+x\right)^4

Lets determine the expansion of the expression

\left(3+x\right)^4

\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i

a=3,\:\:b=x

=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i

Expanding summation

\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}

i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0

i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1

i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2

i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3

i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4

=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4

=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4

as

\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81

\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x

\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2

\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3

\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4

so equation becomes

=81+108x+54x^2+12x^3+x^4

=x^4+12x^3+54x^2+108x+81

Therefore,

  • \left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81
6 0
3 years ago
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