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S_A_V [24]
4 years ago
15

What is the coefficient of x^2 y^3 in the expansion of (2x+y)^5

Mathematics
1 answer:
Anna35 [415]4 years ago
4 0

If it's x²y³ then we know it's the second term of the expansion, that known we can use the combination

C(5, 2) = 5!/(2!.3!) = 10

Then if we had something like

(a + b)^5 our second term would be 10a²b³ but as we can see it's "a²"

And in our case we have 2x as a

So we must do 2² too

2² = 4

10 . 4 = 40

Then our second term of the expansion would be

40x²y³

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Find the value of x which satisfies the following equation.<br> log2(x−1)+log2(x+5)=4
weqwewe [10]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: x = 3

____________________________________

\large \tt Solution  \: :

\qquad \tt \rightarrow \:  log_{2}(x - 1)  log_{2}(x + 5)  = 4

\qquad \tt \rightarrow \:  log_{2} \{(x - 1)(x + 5) \} = 4

[ log (x) + log (y) = log (xy) ]

\qquad \tt \rightarrow \: ( x - 1)(x + 5) =  {2}^{4}

\qquad \tt \rightarrow \:  {x}^{2}  + 5x - x - 5 =  16

\qquad \tt \rightarrow \:  {x}^{2}  + 4x - 5 - 16 = 0

\qquad \tt \rightarrow \:  {x}^{2}  + 4x -21 = 0

\qquad \tt \rightarrow \:  {x}^{2}  + 7x - 3x - 21 = 0

\qquad \tt \rightarrow \:  x(x + 7) - 3(x + 7) = 0

\qquad \tt \rightarrow \: (x + 7)(x - 3) = 0

\qquad \tt \rightarrow \: x =  - 7 \:  \: or \:  \: x = 3

The only possible value of x is 3, since we can't operate logarithm with a negative integer in it.

\qquad \tt \rightarrow \: x = 3

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

4 0
2 years ago
PLEASEEEE PLEASEEEE HELPPPP
ad-work [718]

Answer:

dont understand clearly

Step-by-step explanation:

dont understand clearly

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Answer:

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Step-by-step explanation:

Given

V = πr²h ( isolate r² by dividing both sides by πh )

\frac{V}{\pi h} = r² ( take the square root of both sides )

\sqrt{\frac{V}{\pi h} } = r ← the positive value of r

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anyanavicka [17]

Answer:

5.38516480 or  √ 29

hope this helps

have a good day :)

Step-by-step explanation:

4 0
3 years ago
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