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1320559350 mod 97 =89

There is no brackets so I am not really sure what you want.

1-sec8x/1-sec4x=tan8x/tan4x

Are you asking for all real values of x given that:   ??

I suppose that is what you want.

\(\frac{1-sec(8x)}{1-sec(4x)}=\frac{tan(8x)}{tan(4x)}\\ \)

For questions like 

\(b^2+5b+6\)

you need to look for 2 numbers that multiply to be +6

Since it is + they both must be + or both -

BUT

they have to add to be +5  so they both must be +

ok

we have 

1*6=6   but 1+6 is not 5

2*3=6   and  2+3=5        BINGO!

so we have

\(b^2+5b+6=(b+2)(b+3)\)

Just like Alan said :)))

a²-36  

This is a standard difference of two squares so: a² - 36 → a² - 6² →  (a+6)(a-6)

b²+5b+6  →  (b + 2)(b + 3)

7(x-1)-x(x+1) doesn't factor nicely at all as far as I can see! 

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I assume you want to solve for t.

You need a numerical method.  Simple iteration will do:

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Thanks Alan,

I shall classify it as one of life's mysteries :))

I guess it's just related to the fact that the integral is defined in terms of the limit of a sum of infinitely small segments.

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I've never looked into the theory behind it Melody!  Perhaps I should do so.

Type the determinant function in by hand as follows:

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Then press = or Enter

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But Alan, integral is sum of all values but sigma is only sum of integer values so how can you go from one to the other like that?

That is not worded properly but I am sure you know what I mean ://

The other one:

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Related to calculus as follows:

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2m/.6s

3.33 m/s

I need help calculating tan ^-1 (3/2). An example is tan^-1 (13/10) and x=52

atan(3/2) = 56.30993247402

atan(13/10) = 52.431407971173

I entered it into the web2.0calc as       atan(3/2)=

I have answered enough of these already :/

\(V={r}^{2}*\varpi *h={7,6}^{2}*\varpi *16,3=2958{cm}^{3}\)

I have changed you question so that it makes sense.

use long division to find the quotient A(x) and the remainder  R(x) when P(x) is divided by d(x) and express P(x) in the form d(x) times Q(x) + R(x).

P(x)=x^3+1x^2-3X+85         D(x)=x+5

P(-5)=-125+25+15+85=0   therefore this is not going to be a remainder   :)

\(\begin{array}{rrrrrrrr}\\ &x^2&-4x&+17\\\hline x+5|&x^3&+1x^2&-3x&+85\\ &x^3&+5x^2\\\hline &&-4x^2&-3x&+85\\ &&-4x^2&-20x\\\hline &&&17x&+85\\ &&&17x&+85\\\hline &&&&0 \end{array}\)

\(P(x)=d(x)Q(x)+R(x)\\ where\\ P(x)=x^3+1x^2-3X+85\\ Q(x)=(x^2-4x+17)\\ D(x)=x+5\\ R(x)=0\)

Vietnamese food !!!

Look here

Refer to this 

See my last answer.

Draw a line through c and c'

Draw another line through  D and D'

Where the 2 lines cross will be the centre of the enlargment.

the answer is 56

For how many values of  x is the function     

\(y=\frac{x-2}{x^2-7x+10}\;\;\text{undefined}\)

\(y=\frac{x-2}{(x-2)(x-5)} \)

you cannot divide by zero so x cannot be 2 or 5.   this graph will have holes or asyptotes.

So this is fucntion for all real values of x except for x=2 and x=5

It will be a function in the following domain.

(-infiny,2), (2,5), (5,infity)

cancel down and you get

\(y=\frac{1}{x-5}\\ \)

This is a hyperbola.  The asyptotes are x=5 and y=0 but there will be a hole at (2,-1/3)

here is the graph: