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3 years ago

Item 1 Add. ( x ^2 +1−4 x^ 3 +3 x ^4 )+(− x^ 2 −2 x ^4 + x ^3 +4)

Mathematics

1 answer:

Rzqust [24]3 years ago

8 0

Answer: 5

Step-by-step explanation:

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whats 9+10?idk i cant seem to figure it out heeeeeeeeeeeeeeeeeeeeeeeeeeeelp me pleeeeeeeeeeeeeeeese. it the most hard math probe

KonstantinChe [14]

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3 years ago

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Factorize x^2/16+xy+4y^2

lisov135 [29]

Answer:

$\frac{x^2}{16} +xy+4y^2$ can be factored out as: $(\frac{x}{4} +2\,y)^2$

Step-by-step explanation:

Recall the formula for the perfect square of a binomial :

$(a+b)^2=a^2+2ab+b^2$

Now, let's try to identify the values of $a$ and $b$ in the given trinomial.

Notice that the first term and the last term are perfect squares:

$\frac{x^2}{16} = (\frac{x}{4} )^2\\4y^2=(2y)^2$

so, we can investigate what the middle term would be considering our $a=\frac{x}{4}$ , and $b=2y$ :

$2\,a\,b=2\,(\frac{x}{4}) \,(2\,y)=x\,y$

Therefore, the calculated middle term agrees with the given middle term, so we can conclude that this trinomial is the perfect square of the binomial:

$(\frac{x}{4} +2\,y)^2$

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3 years ago

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4 years ago

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Expand (2x+2)^6<br> How would you find the answer using the binomial theorem?

Yanka [14]

Answer:

Step-by-step explanation:

$\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.$

$\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\$

$(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+ \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.$

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2 years ago

3/4(8p+12) + 3/8(16p-8)

grandymaker [24]

3/4(8p+12)+3/8(16p-8)

=3/4[(8p+12)+1/2(16p-8)]

=3/4[8p+12+8p-4]

=3/4[16p+8]

=3/4*8*[2p+1]

=6[2p+1]

=12p+6

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4 years ago