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dimulka [17.4K]
2 years ago
10

Solve :3(2x+1)^-1 + 5x=-1​

Mathematics
1 answer:
Finger [1]2 years ago
8 0

hope this is ans . Brother

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The sum of four and the product of three and a number x
Ulleksa [173]
4+3x is what it sound like. I hope this helps. :)
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3 years ago
the product of 2 consecutive odd integers is equal to 27 plus six times the sum of the integers. what is the larger integer
olganol [36]
2 consecutive odd integers : x and x + 2

x(x + 2) = 27 + 6(x + x + 2)
x^2 + 2x = 27 + 6(2x + 2)
x^2 + 2x = 27 + 12x + 12
x^2 + 2x - 12x - 12 - 27 = 0
x^2 -10x - 39 = 0
(x + 3)(x - 13) = 0

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x = -3 (extraneous solution)

x - 13 = 0
x = 13

x + 2 = 15 

so ur 2 integers are 13 and 15, with 15 being the largest
8 0
3 years ago
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Harlamova29_29 [7]
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3 years ago
Calculate the length of a triangle if another length is 15 feet and the hypotenuse is 39 feet.
Y_Kistochka [10]
We have to use Pythagorean theorem that says : 

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The missing length of the triangle is : 36 feet
3 0
3 years ago
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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