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  what is the remainder of 1320559350 divided by 97 ?

\(\begin{array}{|rcll|} \hline 1320\ 55\ 93\ 50 \pmod{97} \\\\ 1320 \pmod{97} &=& 59 \\ 59\ 55 \pmod{97} &=& 38 \\ 38\ 93 \pmod{97} &=& 13 \\ 13\ 50 \pmod{97} &=& \color{red}89 \\ \hline \end{array} \)

\(1320\ 55\ 93\ 50 \pmod{97} = \color{red}89 \)

Help

Let \(\angle{TUV} = \varphi\)

\(\begin{array}{|rcll|} \hline \tan(\varphi) &=& \frac{24\cdot \sin(38^{\circ})}{12-24\cdot \cos(38^{\circ})} \\ \tan(\varphi) &=& \frac{24\cdot 0.61566147533}{12-24\cdot 0.78801075361} \\ \tan(\varphi) &=& \frac{14.7758754078}{-6.91225808656} \quad & | \quad \text{II. Quadrant}\\ \tan(\varphi) &=& -2.13763363908 \\ \varphi &=& -64.9294774677^{\circ} + 180^{\circ} \\ \varphi &=& 115.070522532^{\circ} \\ \hline \end{array} \)

TU = ?

\(\begin{array}{|rcll|} \hline \frac{ \sin(38^{\circ}) } {TU} &=& \frac{ \sin(\varphi) } {24} \\ \frac{ \sin(38^{\circ}) } {TU} &=& \frac{ \sin(115.070522532^{\circ}) } {24} \\ TU &=& 24\cdot \frac{\sin(38^{\circ})}{\sin(115.070522532^{\circ}) } \\ TU &=& 24\cdot \frac{0.61566147533}{0.90578692079 } \\ TU &=& 16.3127497965 \\ \hline \end{array} \)

\(TU \approx 16.3\ mm\)

Thank you so much everyone for answering. I'm very grateful.

Can you prove sin4x=(4sinxcosx)(2cos^2x-1)?

\(\begin{array}{|rcll|} \hline \sin(4x) &=& [4\sin(x)\cos(x)][2\cos^2(x)-1] \\ && \cos(2x) = \cos^2(x) - \sin^2(x) = \cos^2(x) - [1- \cos^2(x)] = 2\cos^2(x)-1 \\ \sin(4x) &=& [4\sin(x)\cos(x)][\cos(2x)] \\ && \sin(2x) = 2\cdot \sin(x)\cos(x) \\ \sin(4x) &=& 2\sin(2x)\cos(2x) \\ && \sin(4x)=\sin(2x+2x) = \sin(2x)\cos(2x) + \cos(2x)\sin(2x) = 2\sin(2x)\cos(2x) \\ \sin(4x) &=& \sin(4x) \ \checkmark \\ \hline \end{array}\)

help please!!!

\(\dfrac{2027!+2028!}{2028!-2029!}\)

\(\begin{array}{|rcll|} \hline && \dfrac{2027!+2028!}{2028!-2029!} \\\\ &=& \dfrac {\frac{2028!}{2028} +2028!} {2028!-2028!\cdot 2029} \\\\ &=& \dfrac { 2028!\cdot \left( \frac{1}{2028} + 1\right) } { 2028!\cdot (1-2029) } \\\\ &=& \dfrac { \frac{1}{2028} + 1} { 1-2029 } \\\\ &=& \dfrac { \frac{1}{2028} + 1} { -2028 } \\\\ &=& \dfrac { 1+2028 } { -2028^2 } \\\\ &=& -\dfrac { 2029 } { 2028^2 } \\\\ \hline \end{array} \)

1. USE A CALCULATOR

\(100\tan(50)\)

Answer

\(-27.1901\)

1. Multiply each term in the parenthesis by -4, 3

\(-4(-3x+5)-3x=3(x-4)-6\)

2. Calculate the product

Multiplying two negatives equals a positive: (-)*(-)=(+)

Use rules of multiplication to calculate the expression

\(-4\times (-3x)-4\times 5-3x=3x-12-6\)

3. Collect the like terms by subtracting their coefficients

Extract the negative sign from the expression

\(12x-20-3x=-3x=3x-12-6\)

4. Subtract the numbers

Add the numbers

\((12-3)x-20=3x-(12+6)\)

5. Move variable to the left by adding its opposite to both sides

Move constant to the right by adding its opposite to both sides

\(9x-20=3x-18\)

6. Eliminate the opposites

\(9x-3x-20+20=3x-3x-18+20\)

7. Collect the like terms by subtracting their coefficients

Keep the sign of the number with the larger absolute value and subtract the smaller absolute value from the larger

\(9x-3x=-18+20\)

8. Subtract the numbers

\((9-3)x=+(20-18)\)

9. Divide both sides of the equation by 6

\(6x=2\)

10. Any expression divided by itself equals 1

Write the division as a fraction

\(6x\div 6=2\div 6\)

11. Reduce the fraction with 2

\(x={2\over 6}\)

Answer

\(x={1\over 3}\)

Alternative form

\(x=\approx 0.333333\)

And there goes 30 minutes of my life... 

find the equation of a line, in general form that is parallel to hte line 3x-5y-15=0 and passes threw the point (3,-2)

Formula:

\(\begin{array}{|rcll|} \hline ax+by+c &=& 0 \quad & | \quad \qquad \text{parallel and passes through the point } P\binom{x_y}{y_p} \\\\ ax+by-(ax_p+by_p) &=& 0 \\ \hline \end{array}\)

The equation:

\(\begin{array}{|rcll|} \hline 3x-5y-15 &=& 0\quad & | \quad a=3 \quad b= -5 \\\\ ax+by-(ax_p+by_p) &=& 0 \quad & |\quad x_p = 3 \quad y_p = -2 \\ 3x-5y-[3\cdot 3-5\cdot (-2)] &=& 0 \\ 3x-5y-(9+10) &=& 0 \\ \mathbf{3x-5y-19} & \mathbf{=} & \mathbf{0} \qquad \text{or } \qquad \mathbf{y = \frac35\cdot x -\frac{19}{5} }\\ \hline \end{array} \)

|3x-1|=|2x+6|

\(\begin{array}{|rcll|} \hline |3x-1| &=& |2x+6| \quad & | \quad \text{square both sides} \\ (3x-1)^2 &=& (2x+6)^2 \\ 9x^2-6x+1 &=& 4x^2+24x+36 \\ 9x^2-4x^2-6x-24x+1-36 &=& 0 \\ 5x^2-30x-35 &=& 0 \\ 5x^2-30x-35 &=& 0 \quad & | \quad : 5 \\ x^2-6x-7 &=& 0 \\ (x-7)(x+1) &=& 0 \\ \hline \end{array} \)

x = 7 or x = -1

check:

\(\begin{array}{|rcll|} \hline |3x-1| &=& |2x+6| \quad & | \quad x=-1 \\ |3\cdot(-1)-1| &=& |2\cdot(-1)+6| \\ |-4| &=& |4| \\ 4 &=& 4 \ \checkmark \\\\ |3x-1| &=& |2x+6| \quad & | \quad x=7 \\ |3\cdot7-1| &=& |2\cdot7+6| \\ |20| &=& |20| \\ 20 &=& 20 \ \checkmark \\ \hline \end{array}\)

what is the vertex of this equation 3x^2-11x-4=y

Formula:

\(\begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\\\ x_v &=& -\frac{b}{2a} \\ y_v &=& c - \frac{b^2}{4a} \\ \hline \end{array}\)

Vertex:

\(\begin{array}{|rcll|} \hline y &=& 3x^2-11x-4 \quad & | \qquad a = 3 \qquad b=-11 \qquad c = -4 \\\\ x_v &=& -\frac{(-11)}{2\cdot 3} \\ &=& \frac{11}{6} \\ \mathbf{x_v} &\mathbf{=}& \mathbf{1.8\bar{3}} \\\\ y_v &=& -4 - \frac{(-11)^2}{4\cdot 3} \\ &=& -4 - \frac{121}{12} \\ &=& -\frac{169}{12} \\ \mathbf{y_v} &\mathbf{=}& \mathbf{-14.08\bar{3}} \\ \hline \end{array}\)

3x^2-11x-4=y

We have the form

Ax^2 + Bx + C   = y

The x coordinate of the vertex is given by :  -B/ (2A)  = - (-11) / (2*3)  = 11/6

To find the y coordinate.....put this x value back into the function....and we have

3(11/6)^2 - 11(11/6) - 4

3(121/36) - 121/6 - 4

121/12 - 121/6 - 4

121/12 - 242/12 - 48/12  =

-169/12

So....the vertex is ( 11/6, -169/12 )

I ship it!

[10,000*1.05^N]+[15,000*1.06^N]+[20,000*1.07^N]+[25,000*1.08^N]=100,000

\([10,000*1.05^N]+[15,000*1.06^N]+[20,000*1.07^N]+[25,000*1.08^N]=100,000\\ [1.05^N]+[1.5*1.06^N]+[2*1.07^N]+[2.5*1.08^N]=10\\\)

Answer from Wolfram|Alpha                Just as guest#1 said. :)

Can you give us a specific problem ????

Oh no! I wouldn't worry about that haha, there are lot of math things that I don't understand. :)

Very nice, hectictar.......you're gonna' put me out of a job.....!!!!

Test the points in each option to determine which one fits the data.

Start by testing the first point, x = 1 and y = 3.

A. 3 =? 2(1)

    3 ≠ 2 , so option A is out.

B. 3 =? 4(1²)

    3 ≠ 4 , so option B is out.

C. 3 =? 2(1²) + 1

     3 = 3 , so option C goes to the next round.

D . 3 =? 3(1²)

     3 = 3 , so option D goes to the next round.

Test the second point, x=2 and y = 12.

C. 12 =? 2(2²) + 1

     12 ≠ 9 , so option C is out.

D. 12 =? 3(2²)

     12 = 12 , option D goes to the next round.

Option D is the only remaining choice. Quickly test the other points for D:

27 =? 3(3²)

27 = 27

48 =? 3(4²)

48 = 48

y = 3x² fits all the data.

f(x) = 3x²

With a calculator ...   

Hi Geno,

It is really good to see you again     

I'll give the explanation a go :)

|3x-1|=|2x+6|

+ and +           3x-1 = 2x+6      (1)

+ and -           3x-1 = - ( 2x+2)    (2) 

- and -           -  (3x-1) = - ( 2x+2)    (3) 

- and +           - (3x-1) = + ( 2x+2)    (4) 

If you look at these 4 options you can see that  

1 and 3 are really the same

and

2 and 4 are really the same.   :)

So there are really only 2 distict options :)

Does that explain it good enough for you?

Note, guest that  the  - /-  solution is

- (3x - 1)  = - (2x + 6)

But multiplying through by -1 would produce the first solution that geno presented

And the  -/+   solution is

- (3x - 1)  = (2x + 6)

Again, mutiplying through by -1 produces geno's second solution

So.....these other two are superfluous........

width = 2 inches

length = 3*width = 3*2 = 6 inches

area = length * width

area = 6 inches * 2 inches = 12 square inches