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3 years ago

How do I evaluate this help me please like seriously help

Mathematics

1 answer:

vlabodo [156]3 years ago

3 0

Answer:

8, see below

Step-by-step explanation:

put 13 and 1/2 in equation, then solve

13-.75(12)+8(1/2)

.75×12=9

8×.5=4

13-9+4=8

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Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of x ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As x goes to infinity, each of the 1/x ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

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2 years ago

Giả thiết X là biến ngẫu nhiên X~N(µ, σ^2) thì P( |X-µ| >3,05*σ ) sẽ là

svp [43]

Answer:

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Step-by-step explanation:

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3 years ago

What are the factor pairs of 55 and 5? im on my sisters account.

Aleks04 [339]

Answer:

The factor pairs of 55 and 5 are 1,5,11, and 55

Step-by-step explanation:

55 = 1 x 55 or 5 x 11. Factors of 55: 1, 5, 11, 55. Prime factorization: 55 = 5 x 11.

8 0

2 years ago

Read 2 more answers

HELp I will give brainliest!!

3241004551 [841]

Answer:

$c=\sqrt{65}=8.062.$

Step-by-step explanation:

$\text{Consider } \bigtriangleup DBC, \text{by Pythagoras}\\DB^2+BC^2=DC^2 \\\Rightarrow b^2=DC^2-BC^2\\\Rightarrow b^2= 5^2-3^2=25-9\\\Rightarrow b^2= 16\\\text{Then } , b=4.\\ \text{Consider } \bigtriangleup DBA, \text{by Pythagoras}\\DB^2+BA^2=DA^2\\\Rightarrow c^2=b^2+BA^2\\\Rightarrow c^2=4^2+7^2\\\Rightarrow c^2=16+49\\\Rightarrow c^2=65\\\text{Then } , c=\sqrt{65}=8.062.$

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2 years ago

103.468 round this to the nearest liter

timama [110]

Answer:

103

Step-by-step explanation:

So the .4 in the tenths place is less than 5 it is rounded down/stays the same so its 103

Hope this helps!!

Have a nice day! :D

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2 years ago