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Original investment  =  1350 (pounds)

Interest rate  =  1.04  - 1   = .04   =  4%

Graph "C" is correct.....this is an exponential curve

Crossing over creates more variety by changing the linkage between genes. In a population that does not experience crossing over, the genes on the same chromosome will always be on the same chromosome and therefore inherited together.  People whose chromosomes cross over are more likely to be fans of John Edward.

In populations that do experience crossing over in diplonema during prophase I, the phenotypes will be more diverse because there is a greater chance for genes to be separated from one another and inherited separately. Crossing over does not always happen in the same location, so for populations in which this occurs, there is an exponentially greater number of inheritance combinations.
 

Also, we would expect a larger gene pool and more diversity in the population whose chrosomes frequently cross-over, or much more genetic variability.

Can you confirm that the equation is correct ?

\(\displaystyle 20\cos^{2}x+\sin(4x)=13.\)

n=0;p=0;cycle: a=(1+n);b=int(a/100)%10;c=int(a/10)%10;d=a%10; n++;if(a%9!=0 and b!=9 and c!=9 and d!=9,  goto loop, goto cycle);loop:p=p+1;printa,"  ",;if(n<=810, goto cycle, discard=0;print"Total = ",p

OUTPUT = 648 such numbers.

It seems Huggy is banned. What a crying shame!

Welp, you should give him a huggy ...and a wedgey.   

The equation y = -6t^2 + 51t describes the height (in feet) of a projectile launched from the surface of Mars at 51 feet per second. In how many seconds will the projectile first reach 108 feet in height?

We need to solve this

108 = -6t^2 + 51t       rearrange as

6t^2 - 51t + 108  = 0        divide through by 3

2t^2 - 17t  +  36   = 0       factor

(2t - 9) (t  - 4)    = 0

Set both factors to 0  and solve for t

2t  - 9    = 0                         t  - 4   = 0         

2t   = 9                                t  = 4 sec

t  =9/2  = 4.5 sec

It first reaches 108 ft  at  4 sec

2 second= 51+51=102 feets 

in 1/100 second, 0.51 feet is climbed 

well 108-102 feets = 6 feets

We know that 1/100 second = 0.51 feet 

so in order to find these 6 feets

0.51 *x= 6 

Specific number of seconds is multiplied to find the feets 

x=6/0.51=11.764 approx.=11.8 

2:11.8 (Second, second/100) 

approx

2 seconds and 11.8 sec/100

2.12 seconds!  

I need help solving these please use PEMDAS. ASAP! Thank you.

We have no interest in doing your homework for you.

The asymptote is y = a.

The perimter is 2*sqrt(19) + 2.

John and Jane go rock-climbing together. John climbs a height of (x+5) miles in (x-1) hours and Jane climbs a height of (x+11) miles in (x+1) hours. Is it possible that they were climbing at the same speed? If so, what must have been their speed, in miles per hour?

nvm I got it

That's an understandable mistake.  President Lincoln was referring to the year of the founding of our nation in 1776, which was 87 years earlier than his 1863 speech dedicating a graveyard at Gettysburg Pennsylvania. 

_.

Hello Melody.

\(\begin{array}{|c|l|} \hline n \text{ dices} & \text{probability } \\ \hline 1 & \dfrac{2}{6^1}= \dfrac{2}{6^1} = \mathbf{\dfrac{1}{3}} = \dfrac{\dfrac{6}{3}}{6} = \dfrac{\dfrac{6}{ \dfrac{6}{2} }}{6} = \dfrac{2\cdot 6^{1-1}}{61} \\ \hline 2 & \dfrac{12}{6^2} = \dfrac{12}{36} = \mathbf{\dfrac{1}{3}}= \dfrac{\dfrac{36}{3}}{36}= \dfrac{\dfrac{36}{ \dfrac{6}{2} }}{36} = \dfrac{2\cdot 6^{2-1}}{6^2} \\ \hline 3 & \dfrac{72}{6^3} = \dfrac{72}{216}= \mathbf{\dfrac{1}{3}} = \dfrac{\dfrac{72}{3}}{216}= \dfrac{\dfrac{72}{ \dfrac{6}{2} }}{216} = \dfrac{2\cdot 6^{3-1}}{6^3} \\ \hline \color{red}4 & \dfrac{432}{6^4} = \dfrac{432}{1296}= \mathbf{\dfrac{1}{3}} = \dfrac{\dfrac{432}{3}}{1296}= \dfrac{\dfrac{432}{ \dfrac{6}{2} }}{1296} = \dfrac{2\cdot 6^{4-1}}{6^4} \\ \hline 5 & \dfrac{2592}{6^5} = \dfrac{2592}{7776}= \mathbf{\dfrac{1}{3}} = \dfrac{\dfrac{2592}{3}}{7776}= \dfrac{\dfrac{2592}{ \dfrac{6}{2} }}{7776} = \dfrac{2\cdot 6^{5-1}}{6^5} \\ \hline \ldots & \ldots \\ \hline n & \dfrac{2\cdot 6^{n-1}}{6^n} = \mathbf{\dfrac{1}{3}}= \dfrac{ \phi(6^n) } {6^n} \\ \hline \end{array} \)

Solutions for Triangles

\(\begin{array}{|rcll|} \hline \mathbf{x^2+a^2} &=& \mathbf{(b+c)^2} \\ x^2+a^2 &=& b^2+2bc+c^2 \\ x^2&=& b^2+2bc+c^2-a^2 \\ x &=& \sqrt{b^2+2bc+c^2-a^2} \\ x &=& \sqrt{2bc+ \mathbf{b^2+c^2-a^2}} \\\\ && \boxed{a^2=b^2+c^2-2bc\cos(A)\\ \mathbf{b^2+c^2-a^2 = 2bc\cos(A)} } \\\\ x &=& \sqrt{2bc+ 2bc\cos(A)} \\ x &=& \sqrt{2bc\Big(1+ \mathbf{\cos(A)}\Big)} \\\\ && \boxed{\cos(A)=\cos^2\left(\dfrac{A}{2}\right)-\sin^2\left(\dfrac{A}{2}\right) \\ \cos(A)=\cos^2\left(\dfrac{A}{2}\right)-\left(1-\cos^2\left(\dfrac{A}{2}\right) \right) \\ \mathbf{\cos(A)=2\cos^2\left(\dfrac{A}{2}\right)-1} } \\\\ x &=& \sqrt{2bc\left(1+ 2\cos^2\left(\dfrac{A}{2}\right)-1\right)} \\ x &=& \sqrt{2bc\left( 2\cos^2\left(\dfrac{A}{2}\right) \right)} \\ x &=& \sqrt{4bc \cos^2\left(\dfrac{A}{2}\right) } \\ \mathbf{x} &=& \mathbf{2\sqrt{bc} \cos\left(\dfrac{A}{2}\right)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \sin(\theta) &=& \dfrac{x}{b+c} \quad | \quad \mathbf{x=2\sqrt{bc} \cos\left(\dfrac{A}{2}\right)} \\\\ \mathbf{\sin(\theta)} &=& \mathbf{ \dfrac{2\sqrt{bc} \cos\left(\dfrac{A}{2}\right)}{b+c} } \\\\ \hline \\ \cos(\theta) &=& \dfrac{a}{b+c} \\\\ \mathbf{(b+c)\cos(\theta)} &=& \mathbf{a} \\ \hline \end{array}\)

I am considering this.

5*5=25

4*6=24

3*7=21

2*8=16

1*9=9

Each time the two numbers add to 10.  

Their product seems to be at its biggest when the two numbers are the same. 

The more different the two numbers are, the smaller the product appears to get.

So, I think you should be able to extend this observation in order to answer (or at least attmpt to answer) your question.

Line l_1 has equation 3x - 2y = 1 and goes through A = (-1, -2). Line l_2 has equation y = 1 and meets line l_1 at point B. Line l_3 has positive slope, goes through point A, and meets l_2 at point C. The area of triangle ABC is 3. What is the slope of l_3?

or

simplify n+1 divided by n squared - n - 2 divided by n squared-1

Is this what you want or do you need to put brackets somewhere?

\(\frac{n+1}{n^2}-n-\frac{2}{n^2}-1\)

Square the equation and replace the cos squared term using the identity

\(\displaystyle \cos^{2}(A/2)=(1/2)(1+\cos(A)).\)

Use the cosine rule to replace the cosine term with (b^2 + c^2 - a^2)/(2bc), and simplify.

That should get you to 1 - a^2/((b + c)^2)), so finally cos^2 = 1 - sin^2.

Warum ist das Heureka?  Ich verstehe nicht.   

Ich spreche Deutsch nur einen bisschen.  

Ich übe nur.

Antworten Sie daher bitte hauptsächlich auf Englisch.

Danke