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3 years ago

Find the coefficient of x^2 in the expansion of (2x - 1)^5.

Mathematics

1 answer:

Mashcka [7]3 years ago

5 0

Answer:

So, the coefficient of the x^2 -term is 80

Step-by-step explanation:

so we need to simplify it:

#10(x^2)(2^3) = 10(x^2)(8) = 80x^2

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Answer:

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Step-by-step explanation:

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podryga [215]

Solution :

First term, a₀ = -10 .

Common difference of A.P. , d = -8 - (-10) = 2 .

Last term, aₙ = 24 .

We know, last term is given by :

aₙ = a₀ + (n-1)d

24 = -10 + (n-1)×2

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Now, sum of series is given by :

$S_n = \dfrac{n}{2} \times ( 2a+(n-1)d)\\\\S_n = \dfrac{18}{2} \times ( 2\times -10 + ( 18 -1) \times 2)\\\\S_n = 126$

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3 years ago

Solve the system by substitution.<br> 2.5x-3y=-13<br> 3.25x-y=-14

MrMuchimi

Answer:

x=-4

y=1

Step-by-step explanation:

For the given system of equations,

2.5x-3y=-13.....eqn(1)

3.25x-y=-14.....eqn(2)

We first make y the subject of eqn (2)

This implies that,

y=3.25x+14.....eqn(3)

We substitute eqn (3) into eqn (1)

2.5x-3(3.25x+14)=-13

2.5x-9.75x-42=-13

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Dividing through by 7.25, we obtain

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3.25(-4)-y=-14

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Step-by-step explanation:

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