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3 years ago

What is two plus two

Mathematics

1 answer:

irinina [24]3 years ago

5 0

Four 1+1+1+1=4 1+1=2 2+2=4!!!!!!

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Find the first three terms in the expansion , in ascending power of x , of (2+x)^6 and obtain the coefficient of x^2 in the expa

Nataly_w [17]

Answer:

The first 3 terms in the expansion of $(2 + x)^{6}$ , in ascending power of x are,

$64 , 192 \times x^{1} {\textrm{ and }}240 \times x^{2}$

coefficient of $x^{2}$ in the expansion of $(2+x - x^{2})^{6}$ = (240 - 192) = 48

Step-by-step explanation:

$(2+x)^{6}$

= $\sum_{k=0}^{6}(6_{C_{k}} \times x^{k} \times 2^{6 - k})$

= $6_{C_{0}} \times x^{0} \times 2^{6} + 6_{C_{1}} \times x^{1} \times 2^{5} + 6_{C_{2}} \times x^{2} \times 2^{4}$ + terms involving higher powers of x

= $64 + 192 \times x^{1} + 240 \times x^{2}$ + terms involving higher powers of x

so, the first 3 terms in the expansion of $(2 + x)^{6}$ , in ascending power of x are,

$64 , 192 \times x^{1} {\textrm{ and }}240 \times x^{2}$

Again,

$(2+x - x^{2})^{6}$

= $\sum_{k=0}^{6}(6_{C_{k}} \times (2 + x)^{k} \times (-x^{2})^{6 - k})$

Now, by inspection,

the term $x^{2}$ comes from k =5 and k = 6

for k = 5, the coefficient of $x^{2}$ is , $(-32) \times 6$ = -192

for k = 6 , the coefficient of $x^{2}$ is, $6_{C_{2}} \times 2^{4}$ = 240

so, coefficient of $x^{2}$ in the final expression = (240 - 192) = 48

3 0

3 years ago

Which expression is equivalent to (2^3) ^5?<br> оооо

gregori [183]

Step-by-step explanation:

$\text{Use}\ (a^n)^m=a^{nm}\\\\(2^3)^5=2^{3\cdot5}=2^{15}$

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3 years ago

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Sophie [7]

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3 years ago

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How do I ace a test that I am not prepared for?

Morgarella [4.7K]

You either start studying now or you fake it and guess and get lucky or you cheat

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3 years ago

If probability of catching a cold is 0.03, what is the probability of 3 people out of six catching a cold?

xeze [42]

Answer:

$P(X=x)=0.0004928$

Step-by-step explanation:

-This is a binomial probability problem.

-Given the probability of catching a cold is 0.03, n=6

#The probability of exactly 3 people catching a cold is calculated as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X=3)={6\choose3}0.03^3(1-0.03)^3\\\\=0.0004928$

Hence, the probability of exactly 3 in 6 people catching a cold is 0.0004928

8 0

3 years ago