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To me it does not make sense that you created this question yourself. 

If that were the case why would you not present it in a readable format. 

\( \displaystyle\int_c^d\;\frac{1}{2x \sqrt x}\;dx\\ =\frac{1}{2} \displaystyle\int_c^d\;x ^{(\frac{-3}{2})}\;dx\\ =\frac{1}{2} \displaystyle \left[\;-2x ^{(\frac{-1}{2})}\right]_c^d\\ =-1 \displaystyle \left[\;x ^{(\frac{-1}{2})}\right]_c^d\\ =-1 \displaystyle \left[\;\frac{1}{\sqrt x}\right]_c^d\\ =-1 \displaystyle \left[\;\frac{1}{\sqrt d}-\frac{1}{\sqrt c}\right]\\ =\;\dfrac{1}{\sqrt c}-\dfrac{1}{\sqrt d} \)

\( \displaystyle\int_0^9\;\frac{1}{2x \sqrt x}\;dx\\ =\dfrac{1}{\sqrt 0}-\dfrac{1}{\sqrt 9}\\ =\;undefined\)

\( \displaystyle\int_9^\infty \;\frac{1}{2x \sqrt x}\;dx\\ =\dfrac{1}{\sqrt 9}-\dfrac{1}{\sqrt \infty}\\ =\dfrac{1}{3}\\ \)

I have not checked this so you need to do so. 

The complex numbers \(a\) and \(b\) satisfy \(a\overline{b} = -1 + 5i\).
Find \(\overline{a} b\).

\(\begin{array}{|rcll|} \hline a\overline{b} &=& -1 + 5i \\\\ \overline{a\overline{b}} &=& \overline{-1 + 5i} \quad | \quad \overline{\overline{b}} = b \\\\ \overline{a} b &=& \overline{-1 + 5i} \quad | \quad \overline{-1 + 5i} = -1-5i \\\\ \mathbf{\overline{a} b} &\mathbf{=}& \mathbf{-1-5i} \\ \hline \end{array}\)

Let

\(\omega\)

be a complex number such that

\(\omega^3 = 1\).
Find all possible values of

\(\dfrac{1}{1 + \omega} + \dfrac{1}{1 + \omega^2}\).

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{1 + \omega} + \dfrac{1}{1 + \omega^2}} \\\\ &=& \dfrac{1 + \omega^2+1 + \omega}{(1 + \omega)(1 + \omega^2) } \\\\ &=& \dfrac{2 + \omega + \omega^2 }{1 + \omega^2 + \omega + \omega^3 } \quad | \quad \omega^3 = 1 \\\\ &=& \dfrac{2 + \omega + \omega^2 }{1 + \omega^2 + \omega + 1 } \\\\ &=& \dfrac{2 + \omega + \omega^2 }{2 + \omega + \omega^2} \\\\ &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)

Wait, never mind. There are possibilities. As long as AB-AC is prime, then the area of BCDE is prime. 

Not sure if you know this, but the side lengths of a square are the same. So if the width of the square is 1, the area is 1 which is not prime. I'm pretty sure that this means there is no possibilities.

Try making the expression you're trying to find into one big fraction. Then use Vieta's Formulas and plug coefficients in.

If LaTeX is the only way someone can type on a computer, I find that very weird. Additionally, you should be able to convert it into somewhat readable form. Also, there is a LaTeX function in the Web2.0Calc editor. I don't want to argue about this, just wanted to let you know that you should create a better story next time. Thanks.

Hi, thanks for posting. We will use heron's formula for this, because I assume if you posted on this forum you would have already had googled a triangle area calculator.

Heron's Formula is:

\(A=\sqrt {s(s-a)(s-b)(s-c)}\)

\(s\): The Semiperimeter of a triangle (Half of the perimeter)

\(a,b,c\) = The sides of the triangle.

First Question:

\(A=\sqrt {1.5(1.5-1)(1.5-1)(1.5-1)}\\A=\sqrt {1.5(.5^3)}\\A=\sqrt {0.1875}\\A \approx \boxed{0.433}\)

Second Question:

\(A=\sqrt {50(50-49)(50-33)(50-18)}\\A=\sqrt {50(544)}\\A=\sqrt {27200}\\A \approx \boxed{164.92}\)

Specify if you want a certain formula used, thanks.

Thank you for trying before you post on this forum. I appreciate it. (I am really tired of all the people just being lazy and not trying at all and posting on here).

As for your question.

To find angle \(Q\) and angle \(P\), you need to first solve for \(y\) . We first solve for the unknown angle \(\angle QRP\). We see that this angle and the angle \(\angle QRS\) add up to 180º (Straight Line). As you already did, 180º-150º is 27º.

We know that the sum of the angles in a triangle add up to 180º, so we can now make a linear equation to solve for \(y\). The Equation is as follows:

\((3y+5)+(2y-7)+27=180\)

We then solve for y:

\((3y+5)+(2y-7)+27=180 \)

\( 3y+5+2y-7+27=180\) -- Remove Parentheses

\(5y+25=180\) -- Combine like terms

\(5y=155\) -- Isolate the Variable

\(y=31\)

We can now solve for each of the angles.

\(m\angle Q= 3(31)+5\\m\angle Q= 93+5\)

\(m\angle Q= \) \(\boxed{98}\)

\(m\angle P=2(31)-7\\m\angle P=62-7\\m\angle P=\boxed{55}\)

\(m\angle QRP=\boxed{27}\) (We already solved this.)

This was my first time using LaTeX, please forgive me. :)

1) should be obvious

2)

\(\dfrac{4-\sqrt{18-x}}{x-2} =\\ \dfrac{4-\sqrt{18-x}}{x-2} \dfrac{4+\sqrt{18-x}}{4+\sqrt{18-x}}= \\ \dfrac{16-(18-x)}{(x-2)(4+\sqrt{18-x})} =\\ \dfrac{1}{4+\sqrt{18-x}}\)

now plug x=2 in

AoPS I assume? Or some online class for math? My tips for you. These classes are expensive, and please try to do your homework and at least try to solve the problems before you ask here. If you did indeed attempt to solve these problems, please show your steps to solve. Furthermore, your effort on the post is super minimal, and you don't even format it correctly. At least take the format out of LaTeX and into readable form before you post. We are here to help you learn, not provide answers. If you really don't want to learn at your class/course, please just don't waste your time or sign up for it. Just a message and reminder from me.

I think there's only one possibility.  If the area of that square is a prime number, then either its width or its height has to be 1.  Let's say its width.  Any value for the width other than 1, when multiplied by the height will give a product that isn't prime. 

.

2 or greater

a.  The formula  is   a_n =    a_n-1 - 3....so   a₂ = a_2-1 - 3  =  a₁ - 3

Notice GM  that we are just subtacting 3 from the previous term to get the next....and the first term is 7...so

7, 4, 1 , -2

This is arithmetic

b.  The formula  is  b_n = 2b_n-1  - 1   .....so    b₂ = 2b_(2-1) - 1  =  2b₁ - 1

We are taking  twice the previous term and subtracting 1 to get the next term and the first term is  2

So we have

2, 3, 5, 9

This is neither

c. The formula is  c_n  = 10c_n-1   ...  so.....c₂  = 10 c_{(2 - 1)}  = 10c₁

So....we are just taking 10 times the previous term to h=get the next 

So      3, 30, 300, 3000

geometric

d.   A little tricky.....

We have

 d_n  = n * d_n-1   .....  first term is  1

Second term...n = 2      

d₂ = 2 * d_n-1   =    2 * d_{(2 - 1)}  =  2 * d₁  = 2 * 1  =  2

d₃  =   3* d₂  =  3 * 2  = 6

d₄  = 4 * d₃  = 4 * 6 = 24

So we have

1, 2, 6 , 24

neither......no common difference or common ratio

The common difference is -4

The 14th term is

11 -4(14 - 1)  =

11 - 4(13)

11 - 52 

-41

So.....the sum of the first   14 terms  =

[ first term + last term] * number of terms  / 2    =

[ 11 - 41 ] * 14  / 2  =

[ -30 ] * 7  =

-210

a. 6  12  18  24

Not geometric  because there is not a common ratio between terms

It appears the the common difference is 6

This leads us to believe that  "C" might be correct

Check : 

First term  =  6(1 - 1) +  6  = 6

Second term = 6(2 - 1) + 6 = 6 +6 = 12      yep.....that's the correct one  !!!

b.  2   14   98     686

We have a common ratio of  14 / 2  = 7  [ check GM that  98/14  is the same ratio ]

This is geometric

Pretty sure that "A" is correct

First term = 2 (7) ^(1 - 1)  = 2*7^0  = 2

Second term  2(7)^(2 - 1)  = 2*7^1  = 14

Yep.....that's correct, too

Well....not much mystery to

c. 160  80  40  20

Must be "B"

Both of these have two sides and an included angle

When this is the case......the area will  (1/2) *product of the sides * sine of the included angle

2a.      a = 8, c = 16, B = 60°

Area  = (1/2) ( 8) (16) √3 / 2   =  32√3  units^2

2b.      b = 6, c = 4 sqrt3, A = 30°

Area =  (1/2) (6) (4) (1/2)   =  6 units^2

Just the sum of the integers  1 to 30   =

[ 30] * [ 31 ] / 2 =

15 * 31  =

465 poles

Arithmetic sequence    d = 1

a1 = 30

a30 = 1

S_n= n(a₁+a₃₀)/2 =  30 (30+1)/2 = 465 poles

[0^5 - 0] / 5  = 0/5  = R 0

[1^5 - 1 ] / 5  = 0/5  = R 0

[2^5 - 2] / 5 = 30/ 5  = R 0

[ 3^5 - 3 ] / 5  = 240/5  = R 0

[ 4^5 - 4 ] /5  = 1020 / 5 = R0

Sum =  0

I think I can do part c

3^x

x intercept  - none

y intercept = (0,1)

Increases from left to right

Domain - all reals

Range  - (0, infinity)

Asymptotes  -  y = 0  

End behavior  -  Approaches y = 0 on the left..... Up on the right

log₃ x 

x intercept -  (1,0)

y intercept - none

Increases from left to right

Domain = (0, infinity)

Range = all reals

Asymptotes - x = 0  

End behavior - Down on the left, Up on the right

Not totally sure about this , GM....check the answer

75  =  10 [ log I / I_{0  ]}      divide both sides by 10

75/10   [ log I / 10^(-12) ]   

7.5  = [ log I ] /[ 10^(-12)]      and we can write

7.5 = log I -  log10^(-12)          [ -log 10^(-12)  = -( -12) log 10  =  12 * 1   = 12 ] 

7.5 =   log I  - (-12)

7.5 = log I + 12

7.5 - 12 = log I

-5.5 = log I          using the exponential form

10^(-5.5) w/m^2   =  I