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integrals

1)  \(\displaystyle \int \limits_{0}^{1}\sqrt{1-\sqrt{x}}\;dx\)

 For the integrand  \(\sqrt{1-\sqrt{x}} \) , substitute \(u = \sqrt{1-\sqrt{x}} \) or \(\sqrt{x}=1-u^2\) and

 \(\begin{array}{rcll} du &=& \frac12(1-\sqrt{x})^{\frac12-1} (-\frac12)x^{\frac12-1} \;dx \\ &=& -\dfrac14\cdot \dfrac{1}{\sqrt{1-\sqrt{x}}\sqrt{x}} \;dx \\ &=& -\dfrac14\cdot \dfrac{1}{\sqrt{1-\sqrt{x}}\sqrt{x}} \;dx \\ &=& -\dfrac14\cdot \dfrac{1}{u(1-u^2)} \;dx \\\\ \text{so } dx &=& -4u(1-u^2)\;du \end{array}\)

This gives a new lower bound \(u = \sqrt{1} = 1 \) and upper bound \(u = \sqrt{0} = 0\)
\(\begin{array}{rcll} &=& -4 \displaystyle \int \limits_{1}^{0}u\cdot u(1-u^2)\;du \\ &=& -4 \displaystyle \int \limits_{1}^{0}u^2(1-u^2)\;du \\ &=& -4 \displaystyle \int \limits_{1}^{0}u^2-u^4\;du \\ &=& -4\cdot \Big( \displaystyle \int \limits_{1}^{0}u^2 \;du-\displaystyle \int \limits_{1}^{0} u^4\;du \Big) \\ &=& -4\cdot \Big( \Big[ \frac{u^3}{3} \Big]_{1}^{0} -\Big[ \frac{u^5}{5} \Big]_{1}^{0} \Big) \\ &=& -4\cdot \Big( -\frac{1}{3} - \Big( -\frac{1}{5} \Big) \Big) \\ &=& -4\cdot \Big( \frac{1}{5} -\frac{1}{3} \Big) \\ &=& -4\cdot \left( \dfrac{-2}{15} \right) \\ &=& \dfrac{8}{15} \\ \end{array} \)

Это лучший способ сделать это.

This is a better way to do it.

\(\displaystyle\lim_{x\rightarrow 2}\;\frac{x^2-x-2}{x^2-3x+2}\\ =\displaystyle\lim_{x\rightarrow 2}\;\frac{(x-2)(x+1)}{(x-2)(x-1)}\\ =\displaystyle\lim_{x\rightarrow 2}\;\frac{(x+1)}{(x-1)}\\ =\frac{3}{1}\\ =3\)

7.1

It is important to note that both the numerator and the denominator tend to 0 so I will use L'Hopital's rule.

\(\displaystyle\lim_{x\rightarrow 2}\;\frac{x^2-x-2}{x^2-3x+2}\\ =\displaystyle\lim_{x\rightarrow 2}\;\frac{\frac{d}{dx}(x^2-x-2)}{\frac{d}{dx}(x^2-3x+2)}\qquad \text{Using L'hopital's Rule }\\ =\displaystyle\lim_{x\rightarrow 2}\;\frac{2x-1}{2x-3}\\ =\frac{3}{1}\\ =3\)

The current radius  is   diameter/ 2  = 16/ 2  =  8

And area  = pi * r^2

So....the current area is  pi (8)^2  = 64pi  units^2

Reducing by 48 pi units^2  means that he new area is  [ 64 pi - 48 pi ]  =  16 pi units^2

So......16 pi  = pi (r^2)       .....divide both sides by  pi ......  

16  = r^2  ⇒    4  = r

So....the current radius must be halved to reduce the area by 48 pi units^2

The diameter of a circle is 16. By what number must the radius be decreased in order to decrease the area of the circle by 48 pi?

I really am not sure what you are asking.....

The original area is   pi*64    if you want to decrease this by 48pi  then the new area must be 16pi units^2

So the radius would have to be 4 units, the diameter would be 8 units.

This is half of the orginal diameter. 

Ok thx i figured it out. Thx.

Well....2 pi  meters is the " for sure answer"....since pi is irrational, we won't have a definite "decimal" answer...thus, the approximation.....

Is there a for sure answer? Instead of approxmate.

Thank you so much Hectictar!

1.   y⁴ - 2y² - y³

First we want to look for something that is common to all the terms...we want to take out the GCF . In this case, we can take out  y²  from each term, like this...

=   y²( y² - 2 - y )        Notice how distributing  y²  gets us back to the original expression.

=  y²( y² - y - 2 )

There are three terms remaining...we cannot factor it as a difference of squares.

( A difference of squares is always in the form  a² - b²   ,  just two terms. )

So we need to think of two numbers that add to  -1  and multiply to  -2 .

How about  -2  and  +1  ?    So our expression factors like this...

=  y²( y - 2 )( y + 1 )

2.     x² - xy - x + y        Nothing is common to all the terms. Let's rearrange it like this...

=   x² - x - xy + y           Now we can take out  x  from the first two terms.

=   x(x - 1) - xy + y        Notice that distributing the  x  makes the previous expression.

                                     Now let's take out  -y  from the last two terms.

=   x(x - 1) - y(x - 1)      Again, check this with distrubuting.

Now...there is a common factor in the remaining terms. It is  (x - 1) . Let's factor that out.

=  (x - 1)(x - y)

We can think of it like this :

We have a bike with tires that have a radius of one meter......when the bike moves such that its wheels travel through one revolution.....the linear distance traveled  is just the circumference of the tire =   2 * pi * (1 meter)  ≈  6.28 meters

But......all the parts of the bike will travel the same distance....so....the center of the wheel also travels ≈ 6.28 meters......

We need to solve this :

(p^2 / 2 ) * (p/2)^(n -1)  = p^9 / 256    where n is the nth term

Multiply through by  2 / p^2

(p / 2)^(n -1)  =  p^7 / 128

Note   p^7 / 128  =   p^7 / 2^7  =  ( p / 2)^7

So......

(p / 2) ^(n - 1)  =  (p /2 )^7

Since the bases are the same, solve for the exponents

n - 1  = 7      add 1 to both sides

n  =  8   terms

I think this is correct.....we have a binomial probability...

Probability that a die shows a 1 or 2  = 1/3

Probability that it shows something else  = 2/3

We want to choose  any two of the five die to show a 1 or a 2...so we have.... 

C (5,2) (1/3)^2 (2/3)^3  =  80 / 243   ≈  32.9 %

If the circumference - a linear measure -  increases by 70%..it is now 1.7 times as large........the diameter and the radius which are also linear measures will also be 1.7 times as large.....so this is the scale factor

So......the new area will increase by the square of this scale factor  = (1.7)^2  = 2.89 times = 189%

[ To see this....note that if it doubles....it increases by 100%....and if it triples, it increases by 200%

So.....2.89 times must lie between 100% and 200% ]

This is the "Harmonic Series" raised to an EVEN power, in this case, 14. All these series were extensively studied and proven by the great Swiss mathematician, Leonhard Euler, some 250 years ago!. So, "a" in this sequence is simply π^14.

Welcome to the forum!! 

1.    1 '  =  1/60 °     →     14 '  =  14/60 °

       1 "  =  1/3600 °     →     18 "  =  18/3600 °

136 ° + 14 ' + 18 "  =

136 ° + 14/60 ° + 18/3600 °  =

81743/600 °     ≈     136.238°

2.     1 °  =  60 '     →     0.853 °  =  0.853 * 60 '  =  51.18 '

        1 '   =  60 "     →     0.18 '  =  0.18 * 60 "  =  10.8  "

-( 22.853 ° )  =

-( 22 ° + 0.853 ° )  =

-( 22 ° + 51.18 ' )  =

-( 22 ° + 51 ' + 0.18 ' )  =

-( 22 ° + 51 ' + 10.8 " )  =  - 22 ° 51 ' 10.8 "

3.     cos θ  =  1/3

Use the Pythagorean identity to find  sin θ .

sin² θ  +  cos² θ   =   1

sin² θ  +  (1/3)²  =  1

sin² θ  +  1/9  =  1

sin² θ  =  1 - 1/9

sin² θ  =  8/9

sin θ  =  ±√( 8/9 )    =    ± √8 / 3          Use these values for  sin  and  cos  in the definition of tan .

tan θ  =  sin θ / cos θ

tan θ  =  [ ± √8 / 3 ] / [ 1/3 ]                 Invert the second fraction and multiply.

tan θ  =  [ ± √8 / 3 ] * [ 3/1 ]

tan θ  =  ± √8

tan θ  =  ± 2√2

If you have a question about one of these, please don't hesitate to ask!

Given to us by our Math teacher. Why?

would you mind if I ask where you found this problem?

Maybe we're both wrong, but think that we're correct.....LOL!!!!!

I'm glad you answered it too and got the same thing....I was a little bit unsure of my answer!!!

If

 -c  = 1       multiply both sides by -1

(-1)(- c)   = (-1) 1

c   =    -1

Sorry, hectictar....I didn't see you working on this at the same time !!!

Different procedures....same answer  !!!!

There are 6 choose 3 ways for 3 of the dice to show even numbers and 3 of them to show odd numbers. Each roll is even with probability 1/2 and odd with probability 1/2, so each arrangement of 3 odd numbers and 3 even number occurs with probability (1/2)^6. Thus, the probability that 3 dice out of 6 show even numbers is 5/16

From the AOPS webstite..

x^a  = k   → log x  =  log k / a

y^b  = k   → log y  =  log k / b

x^c  = t   

y^d  = t

So

k * t   =  x^a * x^c =  y^b * y^d  =  x^(a + c) = y^(b + d)

So

(a + c) log x  =  ( b + d) log y

(a + c) ( log k / a)  =  (b + d)( log k / b)

(a + c) b  =  ( b + d) a

ab + bc  =   ab + ad

[ bc   =  ad ]  →  [ ad   =  bc ]

x^a  =  y^b

(x^a)^n  =  (y^b)^n

x^(an)  =  y^(bn)

Let's say

c  =  an     and     d = bn

c/a  =  n     and     d/b  =  n

c/a  =  d/b     cross multiply

ad  =  bc