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 1    2   & 5    are true

S-O-H-C-A-H-T-O-A

sin = opp/hyp     cos = adj/hyp   tan = opp/adj

Integral

\displaystyle \int \limits_0^{1}\frac{\ln(x+1)}{x^2+1}dx

\(\displaystyle \int \limits_0^{1}\frac{\ln(x+1)}{x^2+1}dx\)

\(\begin{array}{|llcl|} \hline \displaystyle \int \limits_{x=0}^{x=1}\dfrac{\ln(x+1)}{x^2+1}\ dx =\ ? \\\\ \text{substitute:}\\ & x = \tan(z) \\ & dx =( \tan^2(z)+1)\ dz \\ \text{new limits:}\\ & x=0 \Rightarrow z = \arctan(0) = 0 \\ & x=1 \Rightarrow z = \arctan(1) = \frac{\pi}{4} \\ \hline \end{array} \)

\(\begin{array}{|rcl|} \hline && \displaystyle \int \limits_{x=0}^{x=1}\dfrac{\ln(x+1)}{x^2+1}\ dx \\ &=& \displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\dfrac{\ln(\tan(z)+1)}{\tan^2(z)+1} ( \tan^2(z)+1)\ dz \\\\ &=& \color{red}\displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(\tan(z)+1) \ dz \\\\ && \boxed{\text{Formula: }\int_0^a f(x)dx=\int_0^a f(a-x)dx} \\\\ &=& \displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln\left(\tan(\frac{\pi}{4}-z)+1 \right) \ dz \\\\ && \boxed{\text{Formula: } \\ \tan(\frac{\pi}{4}-z) = \frac{\tan(\frac{\pi}{4})-\tan(z) }{1+\tan(\frac{\pi}{4})\tan(z)} \\ = \frac{1-\tan(z)}{1+\tan(z)} } \\\\ &=& \displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln\left(\frac{1-\tan(z)}{1+\tan(z)}+1 \right) \ dz \\\\ &=& \displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln\left(\frac{1-\tan(z)+1+\tan(z)}{1+\tan(z)} \right) \ dz \\\\ &=& \displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln\left(\frac{2}{1+\tan(z)} \right) \ dz \\\\ &=& \displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(2)\ dz - \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(1+\tan(z) ) \ dz \\\\ &=& \displaystyle [\ln(2)z]_{z=0}^{z=\frac{\pi}{4}} - \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(1+\tan(z) ) \ dz \\\\ &=& \displaystyle \ln(2)\frac{\pi}{4} - \color{red}\int \limits_{z=0}^{z=\frac{\pi}{4}}\ln (1+\tan(z) ) \ dz \\\\ \color{red}\displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(\tan(z)+1) \ dz &=& \displaystyle \ln(2)\frac{\pi}{4} - \color{red}\int \limits_{z=0}^{z=\frac{\pi}{4}}\ln (1+\tan(z) ) \ dz \\\\ 2\cdot \color{red}\displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(\tan(z)+1) \ dz &=& \displaystyle \ln(2)\frac{\pi}{4} \\\\ \color{red}\displaystyle \int \limits_{z=0}^{z=\frac{\pi}{4}}\ln(\tan(z)+1) \ dz &=& \displaystyle \ln(2)\frac{\pi}{8} \\\\ \displaystyle \int \limits_{x=0}^{x=1}\dfrac{\ln(x+1)}{x^2+1}\ dx &=& \displaystyle \ln(2)\frac{\pi}{8} \\\\ \mathbf{\displaystyle \int \limits_{x=0}^{x=1}\dfrac{\ln(x+1)}{x^2+1}\ dx }& \mathbf{=} & \mathbf{0.27219826129} \\ \hline \end{array}\)

I believe that "losing" or "playing away" would be considered independent events. By definition, "independent events" are events that have no bearing on the future events occurring. In the context of this example, the team could be playing an away game today, but tomorrow the same team could either be playing an away or home game. 

Upon further review, I am not a sports aficionado, so I did not consider the concept of a "home field" advantage. I had, therefore, dismissed the possibility of a dependent event. Thank you, GingerAle, for respectfully correcting me while providing a succinct explanation.  

I have answered on the original thread. That is where you need to go if you want to continue with more questions or answers on this post.

Hi Dot.Box :)

Tertre has given you a very good hint. Thanks Tertre :))

As Tertre said, in similar figures the legths of corresponding sides must be in the same ratio.

So

\(\frac{JK}{MN}=\frac{LJ}{OM}\\ \frac{14}{3.5}=\frac{x}{12}\\\)

Now multiply BOTH sides by 12 to get the value of x.

What answer do you get?

--------

With the second one the two simialar trianges are   LOP and LMP

so

\(\frac{LO}{LM}=\frac{LP}{LN}\)

now do the substitution and solve for x

If you need more help you can ask on THIS thread but make you you say what you have done.

You can certainly do some of it on your own :)

its B. because there is a gap in between the data making it unfair.

Graph the functions on the same coordinate plane.

f(x)=x^2+6x+8

g(x)=−x^2+4

What are the solutions to the equation f(x)=g(x)

so for the graph, part what is the solutions -1 and -2?

For how many integer values of n between 1 and 120 inclusive does the decimal representation of n/120 terminate?

The terminating decimal is if and only if the denominator has powers of only 2 and/or 5

The denominator is \(120 = 2^3\cdot {\color{red}3} \cdot 5^1\)

So \(n\) must drop the 3 in the denominator.

\(n = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, \ldots , 117, 120\)

\(\begin{array}{|lrcll|} \hline AP: & a_n &=& 3+(m-1)*3 \quad & | \quad a_n = 120 \\ & 120 &=& 3+(m-1) * 3 \\ & 120 &=& 3+3m - 3 \\ & 120 &=& 3m \\ & m &=& \frac{120}{3} \\ & \mathbf{m} & \mathbf{=} & \mathbf{40} \\ \hline \end{array}\)

40 integer values of n between 1 and 120 inclusive does the decimal representation of \(\frac{n}{120}\) terminate.

Solve the system of equations.

y=x+5    y=x^2+5x−7

Enter your answers in the boxes.

(  ,  ) or (  ,  )

thankss

A spinner is divided into four equal sections that are numbered 2, 4, 7, and 9. The spinner is spun twice.

How many outcomes have a product less than 30 and contain at least one odd number?

I assume the answer is 6:

\(\begin{array}{|r|r|r|r|} \hline \text{product} & 2 & 4 & 7 & 9 \\ \hline 2 & 4 & 8 & \color{red}14 & \color{red}18 \\ \hline 4 & 8 & 16 & \color{red}28 & 36 \\ \hline 7 & \color{red}14 & \color{red}28 & 49 & 63 \\ \hline 9 & \color{red}18 & 36 & 63 & 81 \\ \hline \end{array}\)

The sum of two positive integers a and b is 80. What is the largest possible value of gcd(a,b)?

\(\begin{array}{|r|r|r|r|} \hline a & b & a+b& gcd(a,b) \\ \hline 1&79&80&1 \\ 2&78&80&2 \\ 3&77&80&1 \\ 4&76&80&4 \\ 5&75&80&5 \\ 6&74&80&2 \\ 7&73&80&1 \\ 8&72&80&8 \\ 9&71&80&1 \\ 10&70&80&10 \\ 11&69&80&1 \\ 12&68&80&4 \\ 13&67&80&1 \\ 14&66&80&2 \\ 15&65&80&5 \\ 16&64&80&16 \\ 17&63&80&1 \\ 18&62&80&2 \\ 19&61&80&1 \\ 20&60&80&20 \\ 21&59&80&1 \\ 22&58&80&2 \\ 23&57&80&1 \\ 24&56&80&8 \\ 25&55&80&5 \\ 26&54&80&2 \\ 27&53&80&1 \\ 28&52&80&4 \\ 29&51&80&1 \\ 30&50&80&10 \\ 31&49&80&1 \\ 32&48&80&16 \\ 33&47&80&1 \\ 34&46&80&2 \\ 35&45&80&5 \\ 36&44&80&4 \\ 37&43&80&1 \\ 38&42&80&2 \\ 39&41&80&1 \\ 40&40&80&\color{red}40 \\ \hline \end{array}\)

The largest possible value of gcd(a,b) is 40

Good point x^2    I was working off of yangg's example and not the question.   my mistake....BUT since the questioner asked about the quadratic formula, I guess we can assume   = 0  

The weight given in this question is irrelavent....not needed....

The surface area of a sphere (bowling ball)   4 pi r^2

You are given DIAMETER...use 1/2 of this for the RADIUS   then   4 pi r^2    in^2

   Can you take it from here?

Remember the equation for a circle with center  (h,k) and radius r is

(x-h)^2 + (y-k)^2 = r^2

so given the info in the question, this equals:

(x-(-4))^2 + (y-3)^2 = 2.5^2

(x+4)^2 + (y-3)^2 = 6.25

Volume of the quarter = pi r^2 x h

     = pi * (15/16 * 1/2)^2 * 1/16

FOUR times this volume is  (the new coin volume)

    4 * pi * (15/16 * 1/2)^2 * 1/16

the new coin volume = pi r^2 h    also

pi * (9/8 * 1/2)^2  h   =   4 * pi * (15/16 * 1/2)^2 *  1/16     solve for the new coin's h

(9/16)^2  h   =  1/4 (15/32)^2

h=  25/144

IF RAM is a diameter, then  ANGLE RAC   is 180 degrees - 40 = 120 degrees.

The TOTAL circumference of the 360 degree circle is  pi x d     =  pi x 2 r = pi x 2 x 8

The fraction of this total circumference is  140 degrees oout of 360 degrees  or   140/360 of the entire circumference

Thi.s equals    (140/360 )  *  pi * 2 * 8 = 19.55 cm

BC*AC=DC*EC

6*10=x*12

60=12x

x=5

DC=5

thank you!

18 degrees.     Central angles are twice inscribed angles with same arc length.

Is the answer around 66%? 

15.75

this is a proportion

lets say that 5.25 to 6 is equal to x to 18. notice that 18 is 3*6 so in order to find x we need to multiply 5.25 by 3 which is 15.75

I am not sure why the answerers have treated an expression as an equation. 

Anyway, as the other answerers have pointed out, the trinomial \(-x^2-14x-24\) is factorable. When factoring, I have one suggestion: Manipulate the expression so that the quadratic term (the x^2-term) is positive. 

....or maybe you can see it more clearly this way:

Multiply both sides of the equation by  -1  to get

x^2+14x+24 = 0

Now factor

(x+2)(x+12) = 0      x = -2 or -12