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I think you want the volume contained between the   7 and 14 cm radii?

Volume of cylinder = pi r^2 h  

Take the volume of the 14 cm cylinder   then subtract the 7 cm cylinder

pi (14^2)*(10)    - pi 7^2(10) = ........   cm ³

1 x 1    + 2x1   + 3x1   +4x1   + 5x1  +6x1   =   21

21 divided by 6 outcomes     21/6 = 3.5                            (I think)

A triangle has coordinates A(0,0), C(1,2), and B(3,1).  Find the perimeter and the area of the triangle.

Hello Guest!

The triangle ABC has the sides a, b, c.
a, b and c are sections of linear functions in the rectangular coordinate system.

Then:

\(A=\int_0^1f_b(x)dx+\int_1^3 f_a(x)dx-\int_0^3 f_c(x)dx\)

\(f_b(x)=2x\\ f_c(x)= \frac{1}{3}x\)

\(f_a(x)=\frac{y_B-y_C}{x_B-x_C}(x-x_C)+y_C\\ f_a(x)=\frac{-1}{2}(x-1)+2=- \frac{1}{2}x+\frac{5}{2}\\ \color{blue} f_a(x)=-\frac{1}{2}x+\frac{5}{2}\\ \)

\({\color{blue}\int_0^1f_b(x)dx}=\int_0^1(2x)dx=| x^2|_0^1=1-0\color{blue}=1\)

\({\color{blue}\int_1^3 f_a(x)dx}=\int_1^3(-\frac{1}{2}x+\frac{5}{2})dx=|-\frac{1}{4}x^2+\frac{5}{2}x|_1^3\\ =(-\frac{9}{4}+\frac{15}{2})-(-\frac{1}{4}+\frac{5}{2})\color{blue}=3\)

\({\color{blue}\int_0^3 f_c(x)dx}=\int_0^3( \frac{1}{3}x)dx=|\frac{1}{6}x^2 |_0^3=\frac{9}{6}-0\color{blue}=\frac{3}{2}\)

\(A=\int_0^1f_b(x)dx+\int_1^3 f_a(x)dx-\int_0^3 f_c(x)dx=1+3-\frac{3}{2}\)

\(\large A=\frac{5}{2}\)

\(P=a+b+c\)

\(a=\sqrt{(2-1)^2+(3-1)^2}=\sqrt{1+4}=\sqrt{5}\\ b=\sqrt{1^2+2^2}=\sqrt{5}\\ c=\sqrt{1^2+3^2}=\sqrt{10}=\sqrt{2}\cdot\sqrt{5}\)

\(\large {\color{blue} P=a+b+c}=\sqrt{2}\cdot\sqrt{5}+2\sqrt{5}=\color{blue}\sqrt{5}\cdot(\sqrt{2}+2)\)

\(\large P=7.6344\)

  !

Sorry, I don't understand your 3rd condition! It seems to conflict with the first? No?

24! = 6.2044840173323943936e+23 = 2^22 * 3^10 * 5^4 * 7^3 * 11^2 * 13 * 17 * 19 * 23

I don't usually do this kind of stuff in C but in Java...

I'd reckon you do something like this.

First, you might want to create an array.

Say,

int input[] = {A1, A2, A3, A4, A5}

Then, compare A1 to A2, use result to compare to A3...

Maybe, use the minToMax function.

I don't really know, go to StackOverflow and search something up.

The missing angle of the first traingle is 180-100-30=50. Since it is a vertical angle, x=180-35-80=65!

LMK if that is correct since I didn't use anything but my head to do this.

x² + x³ + 9 = 8 + x² + x  

x + x² - 8 = x² + 2x - 5        ===>     x - 8   = 2x-5

x² + 9 – x³ = 8 + x – x³       ===>     x² +9 = b +x  

C ) 

3º grau, 1º grau, 2º grau.

A 2015 digit number is multiplied by 54 to yield the last two digits as 68 in the same order. If the same number is multiplied by 46, the last two digits in order are?

Since it has a EVEN exponent    if x   is negative or positive    the  result will be a POSITIVE

   BUT the negative leading coefficint will always turn it back to NEGATIVE

     so as  x  goes  to  inf   OR  - inf     f(x) will always  go to   negative inf

The WHOLE circumference   is    p i x 2r  = 2 pi

  this is a FRACTION of the whole circumference

     pi/3   /   2pi =  1/6

x² + y² = 2x -6y + 6

x² -2x + y² + 6y = 6

x² -2x + 4 + y² + 6y + 9 = 6 + 4 + 9

(x - 2)² + (y + 3)² =  19    

Eqn of a circle with radius² = 19

Circle A, with a radius of 9, has a horizontal chord BC with a length of 10.
From point C, a vertical line extends to point D.
From point D, a horizontal line extends to point E on the circle's circumference.
Line segment DE has a length of 3.
From point E, another vertical line extends to point F on the circle's circumference.
Points B and D connect to form a line segment, and so do points D and F.
What is the angle (in degrees) of \(\angle BDF\)?

\(\begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle AEN:} \\ \hline \dfrac{y^2}{4} + 8^2 &=& 9^2 \\\\ \dfrac{y^2}{4} &=& 9^2- + 8^2 \\\\ \dfrac{y^2}{4} &=& 17 \\\\ \mathbf{y^2} &=& \mathbf{68} \\ \mathbf{y} &=& \mathbf{2\sqrt{17}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle DEF:} \\ \hline v^2 &=& 3^2 + y^2 \\ v^2 &=& 9+68 \\ \mathbf{v^2} &=& \mathbf{77} \\ \mathbf{v} &=& \mathbf{\sqrt{77}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle BMA:} \\ \hline 5^2+z^2 &=& 9^2 \\ z^2 &=& 81-25 \\ z^2 &=& 56 \\ \mathbf{z} &=& \mathbf{2\sqrt{14}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x + \dfrac{y}{2} &=& z \\\\ x &=& z- \dfrac{y}{2} \\\\ x &=& 2\sqrt{14}- \dfrac{2\sqrt{17}}{2} \\\\ \mathbf{x} &=& \mathbf{2\sqrt{14}- \sqrt{17}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle BCD:} \\ \hline u^2 &=& 10^2+x^2 \\ u^2 &=& 100+ \left( 2\sqrt{14}- \sqrt{17} \right)^2 \\ u^2 &=& 100+ 4*14-4\sqrt{14*17} + 17 \\ \mathbf{u^2} &=& \mathbf{173-4\sqrt{238}} \\ \mathbf{u} &=& \mathbf{\sqrt{173-4\sqrt{238}}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle BPF:} \\ \hline w^2 &=& 13^2 +(x+y)^2 \\ w^2 &=& 13^2 +\left( 2\sqrt{14}- \sqrt{17}+2\sqrt{17} \right)^2 \\ w^2 &=& 13^2 +\left( 2\sqrt{14}+\sqrt{17} \right)^2 \\ w^2 &=& 13^2 +4*14+4\sqrt{14*17} + 17 \\ \mathbf{w^2} &=& \mathbf{242+4\sqrt{238}} \\ \hline \end{array}\)

cos-rule:

\(\begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle BDF:} \\ \hline \mathbf{w^2} &=& \mathbf{u^2+v^2-2uv\cos(\theta)} \\\\ \cos(\theta) &=& \dfrac{u^2+v^2-w^2}{2uv} \\\\ \cos(\theta) &=& \dfrac{173-4\sqrt{238}+77-(242+4\sqrt{238})}{2\sqrt{173-4\sqrt{238}}\sqrt{77}} \\\\ \cos(\theta) &=& \dfrac{8-8\sqrt{238}} {2\sqrt{13321-308\sqrt{238}}} \\\\ \cos(\theta) &=& \dfrac{4(1-\sqrt{238})} {\sqrt{13321-308\sqrt{238}}} \\\\ \cos(\theta) &=& \dfrac{-57.7089944822} {92.5710938948} \\\\ \cos(\theta) &=& -0.62340188556 \\ \theta &=& \arccos(-0.62340188556) \\ \mathbf{\theta} &=& \mathbf{128.564985997^\circ} \\ \hline \end{array}\)

The angle (in degrees) of \(\angle BDF\) is \(\mathbf{128.564985997^\circ}\)

Notice that the curve goes through the following three points (you could choose any three, but these are easy ones):

 (0, 5)   (2, 1)  and  (4, 5)

This allows you to set up the following three equations:

5 = a*0²  +  b*0 + c     or simply  c = 5

1 = a*1² + b*1 + c     or   a + b = -4         ...(1)

5 = a*4² + b*4 + c    or    16a + 4b = 0    ...(2)

You could solve equations (1) and (2) for a and b if you wish; however, since you are only asked for a + b + c you could simply use the value of a + b from (1) together with the value for c.

Thank you so much for your answer to my first problem!

Can anyone help me with the second?

That is incorrect.

yes  , Im sorry , I dont know English math language

That is incorrrect.

Anybody?

Anyone?

This question seems to occur frequently!  Have a look here: https://web2.0calc.com/questions/i-m-really-stumped-polynomials 

I need help with this system:
\(\begin{eqnarray} x(y+z)&=&39\\ y(x+z)&=&60\\ z(x+y)&=&63 \\ x^2+y^2+z^2&=& \ ? \end{eqnarray}\)

\(\begin{array}{|lrcll|} \hline & x(y+z)&=&39 \\ & xy+xz &=& 39 \\ & \mathbf{xz} &=& \mathbf{39-xy} \qquad (1) \\\\ & y(x+z)&=&60 \\ & yx+yz &=&60 \\ & \mathbf{yz} &=& \mathbf{60-xy} \qquad (2) \\ \hline (1)+(2): & xz+yz &=& 39-xy+60-xy \\ & z(x+y) &=& 99-2xy \quad | \quad \mathbf{z(x+y)=63} \\ & 63 &=& 99-2xy \\ & 2xy &=& 99-63 \\ & 2xy &=& 36 \\ & \mathbf{xy} &=& \mathbf{18} \quad \text{or} \quad \mathbf{y=\dfrac{18}{x}} \qquad (3) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline x(y+z)&=& 39 \quad | \quad \mathbf{y=\dfrac{18}{x}} \\\\ x\left(\dfrac{18}{x}+z\right) &=& 39 \\\\ 18 +xz &=& 39 \\\\ xz &=& 39-18 \\\\ \mathbf{xz} &=& \mathbf{21} \quad \text{or} \quad \mathbf{x=\dfrac{21}{z}} \qquad (4) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline z(x+y)&=& 63 \quad | \quad \mathbf{x=\dfrac{21}{z}} \\\\ z\left(\dfrac{21}{z}+y \right)&=& 63 \\\\ 21+zy &=& 63 \\\\ zy &=& 63-21 \\\\ \mathbf{zy} &=& \mathbf{42} \qquad (5) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \dfrac{(3)}{(5)}: & \dfrac{xy}{zy} &=& \dfrac{18}{42} \\\\ & \dfrac{x}{z} &=& \dfrac{3}{7} \quad | \quad \mathbf{x=\dfrac{21}{z}}\quad \text{or} \quad \mathbf{z=\dfrac{21}{x}} \\\\ & \dfrac{x}{\dfrac{21}{x}} &=& \dfrac{3}{7} \\\\ & \dfrac{x^2}{21} &=& \dfrac{3}{7} \\\\ & x^2 &=& \dfrac{3*21}{7} \\\\ & x^2 &=& 3*3 \\\\ & \mathbf{x} &=& \pm \mathbf{3} \\ \hline & \mathbf{y} &=& \mathbf{\dfrac{18}{x}} \\\\ & y &=& \dfrac{18}{\pm 3} \\\\ & \mathbf{y} &=& \mathbf{\pm 6 } \\ \hline & \mathbf{z} &=& \mathbf{\dfrac{21}{x} } \\\\ & z &=& \dfrac{21}{\pm 3} \\\\ & \mathbf{z} &=& \mathbf{\pm 7} \\ \hline & x^2+y^2+z^2 &=& (\pm 3)^2+(\pm 6)^2+(\pm 7)^2 \\\\ & &=& 9+36+49 \\\\ &\mathbf{ x^2+y^2+z^2} &=& \mathbf{94} \\ \hline \end{array}\)