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Ahh thank you very much! 

Sorry for the mistake the correct answer is actually 2, i made a mistake during the ''solve for a'' part. So instead of \(2\sqrt{2}\) the correct answer is \(2\)

Which is oppistie which is adjacent and which is the hyptounuse

Mark them like this O for oppisite A for adjacent and H for hypotonuse.

46•2=92

Well, there are 94.2 inches of rubber around the wheel, which means that the circumference is 94.2. Circumference is pi (3.14) times the diameter, so we get 94.2=3.14•(diameter). If we divide both sides by 3.14, we get the diameter to be 30, and the radius to be 15. The area of a circle is 3.14 times the radius squared, so that gives us 3.14•225, which is 706.5 square inches. Lastly, if you want to cut the circles out of a rectangle, you need to take the diameter of one circle as the height and the diameter of four circles as the width. So, the rectangle needs to be at least 30 inches tall and at least 120 inches wide. "It should probably be a little longer" doesn't add any exact measurements, but it probably means to add a few inches of felt around each of the circles.

Hmm.  You might want to check your analysis Misaki.  I get the following:

.

Solved. 

Just confused by one formula part.

I'm doing good this morning 

calculate the area bounded by the graphs of f(x)=x2 , g(x)=1/x2, x>0 and the line x=3    :)

1. cut\( f(x) \text{ and } g(x) , x \gt 0\)

\(\begin{array}{rcll} f(x) &=& g(x)\\ x^2 &=& \frac{1}{x^2} \\ x^4 &=& 1 \\ x &=& 1 \end{array}\)

2. calculate the area

\(\begin{array}{rcll} \int \limits_{1}^{3} { ( f(x)-g(x) )\ dx } &=& \int \limits_{1}^{3} { f(x) \ dx } -\int \limits_{1}^{3} { g(x) \ dx } \\ &=& \int \limits_{1}^{3} { x^2 \ dx } -\int \limits_{1}^{3} { \frac{1}{x^2} \ dx } \\ &=& [\frac{x^3}{3}]_{1}^{3} - [-\frac{1}{x} ]_{1}^{3} \\ &=& \frac13\cdot [x^3]_{1}^{3} + [ \frac{1}{x} ]_{1}^{3} \\ &=& \frac13\cdot (3^3-1^3) + ( \frac{1}{3}-\frac{1}{1} )\\ &=& \frac13\cdot (27-1) + ( -\frac{2}{3} )\\ &=& \frac13\cdot (26) -\frac{2}{3} \\ &=& \frac{26-2}{3} \\ &=& \frac{24}{3} \\ &=& \frac{3\cdot 8}{3} \\ \mathbf{ \int \limits_{1}^{3} { ( f(x)-g(x) )\ dx } } & \mathbf{=} & \mathbf{8}\\ \end{array}\)

in the number sequence 1,1,2,3,5,8,13.... What is the remainder when the 1995th number is divided by 8? 

\(f_{1995} =\)

\(f_{1995} \pmod 8 = 2\)

First solve the indefinite integral

\(\int 4-2x dx=4x-x^2+c\)

Find boundaries 

\(\int_{0}^{a}4-2xd=-(a-4)a-0\)

Simplify 

\(-(a-4)a\)

\(-(a-4)a=4 \)Solve for a

\(a^2=8\)

\(a=\sqrt{8}\)

\(a=2\sqrt{2}\)

The grid is where?

Use binet's formula to find the 1995th term of the series hint: it's the fibonnacia series

so \(F(n)=\frac{\phi^n-(-\phi)^n}{\sqrt5}\)Let n=1995

\(F(1995)=\frac{\phi^{1995}-(-\phi)^{1995}}{\sqrt{5}}\)

Let Phi=1.6 \(F(1995)=23668326.9616766279790584\) divided by 8 it equals

\(=2958540.8702095784973823\)

Please give the next 5 terms of this sequence and explain your conclusions:

1, 4, 6, 9, 12, 16, 21, 28, 38, 53....... Thanks to all and have fun.

\(\small{ \begin{array}{|l|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline &1&4&6&9&12&16&21&28&38&53&{\color{grey}76}&{\color{grey}112}&{\color{grey}169}&{\color{grey}260}&{\color{grey}406} \\ \hline \text{Fibonacci numbers}&0&1&1&2&3&5&8&13&21&34&55&89&144&233&377 \\ \hline \text{odd numbering}&1&3&5&7&9&11&13&15&17&19&21&23&25&27&29\\ \hline \text{add }& \\ \hline \end{array} }\)

Oh ok thanks!