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huh im confused and its due today

The perpendicular bisector of the line segment connecting the points (-3,8) and (-5,4) has an equation of the form y=mx+b. Find m+b.

Hello xplosions!

Die senkrechte Halbierende des Liniensegments, das die Punkte (-3,8) und (-5,4) verbindet, hat eine Gleichung der Form y = mx + b. Finde m + b.

Two point equation

\(f_1(x)=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1\)

\(f_1(x)=\frac{4-8}{(-5)-(-3)}(x-(-3))+8\\ f_1(x)=\frac{-4}{-2}(x+3)+8\)

\(f_1(x)=2x+14\)

\(m_1=2\\ b_1=14\)

\(P_2(\frac{-3+(-5)}{2},\frac{8+4}{2})\)

\(P_2(-4,6)\)

\(m_2=-\frac{1}{m_1}\)

\( m_2=-\frac{1}{2}\)

Point direction equation

\(f_2(x)=m_2(x-x_2)+y_2\)

\(f_2(x)=-\frac{1}{2}(x-(-4))+6\)

\(f_2(x)=-\frac{1}{2}x+4 \)

\(b_2=4\)

\(m_2+b_2=\frac{7}{2}\)            Wozu braucht man diese Addition?

  !

Let the price of 1 apple =A

Let the price of 1 banana = B

4A  +  6B = 1.56..................(1)

9A  +  7B =2.60..................(2)

Now, you have 2 simultaneous equations. Can you solve them?

Your number is =10,989

10,989 x 9 =98,901

Let

 \(\displaystyle S=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\frac{4}{2^{4}}+\dots +\frac{k}{2^{k}}+\dots \\=S_{1}+S_{2}+S_{3}+\dots+ S_{k}+\dots,\)

where

\(\displaystyle S_{1}=\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\dots, \)

\(\displaystyle S_{2}=\frac{1}{2^{2}}+\frac{1}{2^{3}}+\frac{1}{2^{4}}+\dots.\)

\(\displaystyle S_{3}=\frac{1}{2^{3}}+\frac{1}{2^{4}}+\frac{1}{2^{5}}+\dots,\)

.

\(\displaystyle S_{k}=\frac{1}{2^{k}}+\frac{1}{2^{k+1}}+\frac{1}{2^{k+2}}+\dots,\)

.

.

Each   

\(\displaystyle S_{1},S_{2},S_{3}\dots,\)

is an infinite GP, (sum to infinity = a/(1 - r)) ) with common ratio 1/2. so,

\(\displaystyle S=\frac{1/2}{1-1/2}+\frac{1/2^{2}}{1-1/2}+\frac{1/2^{3}}{1-1/2}+\dots,\\ \displaystyle = 1+\frac{1}{2}+\frac{1}{4}+\dots,\)

which is again a GP common ratio 1/2.

\(\displaystyle S = 2.\)

yes thank you verymuch

As follows:

im not understand

um i dont understand but thankyou

Find the value of the sum \(\dbinom{99}{0} - \dbinom{99}{2} + \dbinom{99}{4}- \dbinom{99}{6} + \dots - \dbinom{99}{98}\)

\((1+i)^{99}=\dbinom{99}{0}+i\dbinom{99}{1}-\dbinom{99}{2}-i\dbinom{99}{3}+\dbinom{99}{4}+i\dbinom{99}{5}- \ldots -\dbinom{99}{98}-i\dbinom{99}{99} \qquad (1)\)

Note here that \(\binom{99}{0} - \binom{99}{2} + \binom{99}{4}- \binom{99}{6} + \dots - \binom{99}{98}\) is the real part of (1).

\(\begin{array}{|rcll|} \hline 1+i &=& \sqrt{1^2+1^2} \Bigg(\cos\Big( \arctan(\frac{1}{1} )\Big) +i\sin\Big( \arctan(\frac{1}{1} )\Big) \Bigg) \\ 1+i &=& \sqrt{2} \Big(\cos(45^\circ) +i\sin(45^\circ) \Big) \\ \hline (1+i)^{99} &=& \Bigg( \sqrt{2} \Big(\cos(45^\circ) +i\sin(45^\circ) \Big) \Bigg)^{99} \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(\cos(99*45^\circ) +i\sin(99*45^\circ) \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(\cos(4455^\circ) +i\sin(4455^\circ) \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(\cos(135^\circ) +i\sin(135^\circ) \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(-\dfrac{\sqrt{2}}{2} +i\dfrac{\sqrt{2}}{2} \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(-2^{\frac{1}{2}-1} +i\dfrac{\sqrt{2}}{2} \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}} \Big(-2^{-\frac{1}{2}} +i\dfrac{\sqrt{2}}{2} \Big) \\ (1+i)^{99} &=& 2^{\frac{99}{2}}*(-2^{-\frac{1}{2}}) +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ (1+i)^{99} &=& -2^{\frac{99}{2}}2^{-\frac{1}{2}} +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ (1+i)^{99} &=& -2^{\frac{99}{2}-\frac{1}{2}} +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ (1+i)^{99} &=& -2^{\frac{98}{2}} +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ (1+i)^{99} &=& \underbrace{\color{red}-2^{49}}_{\text{the real part}} +i*2^{\frac{99}{2}} *\dfrac{\sqrt{2}}{2} \\ \hline \end{array} \)

\(\dbinom{99}{0} - \dbinom{99}{2} + \dbinom{99}{4}- \dbinom{99}{6} + \dots - \dbinom{99}{98} = \mathbf{-2^{49}}\)

I'll include the question because I suspect it will be edited.

A= 26 - x * 26 - y

Answer:

A=26-26x-y

Yes I see, you are correct, it is not right.  the correct answer is equally simple though.

The angles in a right triangle form an arithmetic progression.  If the smallest angle is 13 degrees, then what is the largest angle?

I think that nither arithmetic nor geometric progression can be applied to the angles of the right triangle.

 arithmetic progression       13º         51.5º       90º       wrong!

 geometric progression        13º            34.205275º            90º    

N mod 2 = 1

N mod 3 = 1

N mod 4 = 1

N mod 5 = 1

N mod 6 = 1

N mod 7 =0

i=0;j=0;m=0;t=0;a=( 2, 3, 4, 5, 6, 7);r= (1, 1, 1, 1, 1, 0);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

420m + 301, where m=0, 1, 2, 3......etc.

The minimum number of gold coins the husband sent was 301 coins.

This question is too easy. Try this one: find the area of a rectangle if its area is 10 times the area of a smaller circle.

Let a < b < c < d be four consecutive positive integers such that     

- a is divisible by 5,     
- b is divisible by 7,    
- c is divisible by 9, and      
- d is divisible by 11. 

Find the minimum value of a+b+c+d. 

\(\begin{array}{ll} \left.\begin{array}{rcl} a & \equiv & 0 \pmod{5} \\ b=a+1 & \equiv & 0 \pmod{7} \\ c=a+2 & \equiv & 0 \pmod{9} \\ d=a+3 & \equiv & 0 \pmod{11} \end{array}\right\} \begin{array}{rcl} a & \equiv & 0 \pmod{5} \\ a & \equiv & -1 \pmod{7} \\ a & \equiv & -2 \pmod{9} \\ a & \equiv & -3 \pmod{11} \end{array} \\\\ a+b+c+d = 4a+6 \\ \end{array} \)

\(\begin{array}{ll} \left.\begin{array}{rcl} a & \equiv & 0 \pmod{5} \\ a & \equiv & -1 \pmod{7} \\ \end{array}\right\} \begin{array}{rcl} a & \equiv & -15 \pmod{35} \\ \end{array} \\ \left.\begin{array}{rcl} a & \equiv & -2 \pmod{9} \\ a & \equiv & -3 \pmod{11} \end{array}\right\} \begin{array}{rcl} a & \equiv & -47 \pmod{99} \\ \end{array} \end{array} \)

\(\begin{array}{ll} \left.\begin{array}{rcl} a & \equiv & -15 \pmod{35} \\ a & \equiv & -47 \pmod{99} \\ \end{array}\right\} \begin{array}{rcl} a & \equiv & 1735 \pmod{3465} \\ a &=& 1735 + 3465n,\ n\in \mathbb{Z} \\ \end{array} \\ \end{array} \)

\(\text{The minimum value of $\mathbf{a}$ is $\mathbf{1735}$} \\ \text{The minimum value of $\mathbf{a+b+c+d} = 6*1735 + 4 \mathbf{=6946}$ } \)

Check:

\(\begin{array}{|rcll|} \hline 1735 & \equiv & 0 \pmod{5} \\ 1736 & \equiv & 0 \pmod{7} \\ 1737 & \equiv & 0 \pmod{9} \\ 1738 & \equiv & 0 \pmod{11} \\ \hline \end{array}\)

triangle side     a =  √ [ A / (√ 3)/ 4)]

circle radius     r = [a/2 * tan(30º)] * 2

rectangle area    A = a * r    

wow I really wish I thought of that !!! thanks 

In the cyclic quadrilateral WXYZ on the circle centered at O, ∠ZYW=10º and ∠YOW=100º.  What is the measure of ∠YWZ?

When the quadrilateral is inscribed in a circle, the sum of the opposite angles is 180º.

If ∠YOW=100º then ∠WXY = 50º

If ∠WXY = 50º then the opposite angle (∠YWZ) must be 130º.  

There are three consecutive multiples of five. The sum of the first two numbers is 1,945.
What are the integers?

\(\begin{array}{rcll} x~ \text{is the multiply of}~ 5 \\ \text{First} &=& x \\ 2\text{nd} &=& x+5 \\ 3\text{rd} &=& (x+5)+5 \\ \end{array} \)

\(\begin{array}{|rcll|} \hline x+x+5 &=& 1945 \\ 2x+5 &=& 1945 \\ 2x &=& 1945-5 \\ 2x &=& 1940 \\ \mathbf{x} &=& \mathbf{970} \\ \hline \end{array}\)

\(\begin{array}{rcll} \text{First} &=& 970 \\ 2\text{nd} &=& 970+5 \\ &=& 975 \\ 3\text{rd} &=& 975+5 \\ &=& 980 \\ \end{array}\)

c = 14.

You should try using induction.