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Sin(95)/12 =sin(ADC) / 8

ADC =41.616 degrees

ACD = 180 -95 -41.616 =43.384 degrees

Sin(95) /12 =sin(43.384) /AD

AD =8.274 cm

Area of ACD =32.971 cm^2

Area of Parallelogram ABCD =32.971 x 2 =65.942 cm^2

26ft if i think i might be wrong

thanks dark

Since CD bisects angle C, then:

2.25 / 3 = x / 4        cross multiply

3x = 4 x 2.25           divide both sides by 3

x = [4 x 2.25] / 3

x = 9 / 3

x = 3

6 / 36 =x / 42         cross multiply

36x = 42 x 6          divide both sides by 36

x = [42 x 6] / 36

x = 252 / 36

x = 7

hope your familys ok 

A*105 + 104 =B*111 + 110=C*121 + 111 =D*122 + 111

By simple iteration A=124,985, B = 118,229, C = 108,458, D = 107,569

N = 105 x 124,985 +104 =13,123,529 - the smallest positive integer.

Since the LCM of {105, 111, 121, 122} =57,350,370, therefore:

N = 57,350,370 n + 13,123,529, where n =0, 1, 2, 3...........etc.

yesterday at 1:15 PM my time and 2:15 their time

When was that?

if you make a graph you can find the y-intercept but you need a point like a really good point. 

Janis is 26, she pent 1/2 her life in school => 13yrs

Sorry its not 10

FG=10

Perhaps someone else would like to help with this explanation...??

Any takers?

Here is a different example:

let

$f(x)=x^2$

This is a concave up parabola with vertex (0,0)

Now $f(2x) = (2x)^2\\f(2x) = 4x^2$

How does the second one differ from the first.

Here is the graphs

The green one is f(x) and the blue one is f(2x)

Can you see that for any given y value the new corresponding x is half the size.

So when y=36 the x value was originally $\pm6$     $If \;f(x)=x^2 \;\;then \;\;f(6)=36$

But when x is replaced with 2x then  it becomes   $f(2x) =4x^2$  as show above and  $f(3)=36$

so if you double the x in the equation then the resulting graph will have half the width!

3) Def of Bisector

It is already given info that $\overline{BD}$ bisects $\angle ABC$, so you can use the definition of a bisector (which states that a bisector divides a figure into two congruent parts) to make the conclusion.

5) Alternate Interior Angles Theorem

Both $\angle ABD$ and $\angle BAE$ have a relationship because the angles are formed on opposite side of the transversal, $\overline{BA}$, and both lie in the inner region of the parallel segments. This indictates that both angles are alternate interior angles. Since parallel lines and transversals exist in this scenario, it is possible to use the above theorem to conclude the angles' relationship with each other. 

9) Substitution Property (of Equality)

On line 7, it was established that $AB=EB$. On line 8, it was established that $\frac{AD}{DC}=\frac{EB}{BC}$. If $AB=EB$, then we can use the Substitution Property of Equality to conclude that $\frac{AD}{DC}=\frac{AB}{BC}$, which is identical to the conclusion made. 

$p^4-1$ is a difference of squares. To make this clearer, the original expression can be rewritten as $\left(p^2\right)^2-1^2$. Since the original expression is a difference of squares, it is possible to factor it via $(p^2+1)(p^2-1)$. Notice that one of the factors of the original expression is also a difference of squares, namely $p^2-1$, so it is possible to factor even further. Therefore, the furthest factored expression would be $(p^2+1)(p+1)(p-1)$

Hi all,

Supermanaccz has asked me to come back an talk about this ancient question.

Thanks for your interest Supermanaccz    

Firstly the question is not displaying properly because the coding used on this site has changed a lot in the past 3 and a bit years.

So I will try to put the question up how it may originally have been. I believe a diagram like the one below would have been included in the original question.  Thie is the graph of   y=f(x).

I have drawn it in a program called GeoGebra. GeoGebra is a great tool. I found it a little hard to use at first (especially for more complicated things) but if you persist with it it will become much easier. Geogebra is a free download. I use it often.

Now to the actual question:

The graph of  is shown below. Assume the domain of  is  [-4,4] and that the vertical spacing of grid lines is the same as the horizontal spacing of grid lines. 

Part (a): The points  (a, 4) and  (b, -4)  are on the graph of  y=f(2x)    Find  a  and  b

From the graph you can see that 2 points on the f(x)  graph are (-4,4) and (4,-4)

-------------------------------------

Here is what i said before:

Part a       y=f(2a)

f(-4)=4   but    f(2a)=-4    so   2a=-4     so   a=-2

f(4)=-4   but    f(2b)=4     so    2b=4      so    b=2

----------------------------------------

But in the next post I will do an entirley different example, perhaps you will be able to see how many graphs are related to parent graphs. In this case f(x) is the parent function and f(2x) it the 'offspring'  Multiplying the x by 2 will cause the graph to have half the width (from the y axis). 

Part (b): Find the graph of  y=2x         Verify that your points from part (a) are on the graph. 

Part (c): The points (c,4)  and (d,-4)  are on the graph of y=f(2x-8)     Find  c   and  d

Part (d): Find the graph of  y=f(2x-8)   Be sure to verify that your points from part (c) are on the graph both algebraically and geometrically. 

What is the smallest positive integer that will satisfy the following congruences:

N mod 105 = 104,

N mod 111 = 110,

N mod 121 = 111,

N mod 122 = 111

Thanks for any help.

$\begin{array}{|l|cll|} \hline \begin{array}{|rcll|} \hline \mathbf{ n } & \mathbf{\equiv} & \mathbf{104 \pmod{105}} \\ \mathbf{ n } & \mathbf{\equiv} & \mathbf{110 \pmod{111}} \\ \hline \end{array} \\ \begin{array}{lrcll} \Rightarrow & n -104 &=& 105x \\ & n -110 &=& 111y \\\\ \Rightarrow & n &=& 104+105x \\ & n &=& 110+111y \\\\ & n=104+105x&=& 110+111y \\ & 104+105x&=& 110+111y \\ & 105x-111y&=& 6 \quad & | \quad : 3 \\ & \mathbf{35x-37y}& \mathbf{=}& \mathbf{2} \qquad x,y \in Z \\ &\rightarrow \quad x &=& -1+37b^{^1)} \qquad b \in Z \\\\ & n &=& 104+105x \\ & n &=& 104+105(-1+37b) \\ & n &=& 104-105+105\cdot 37b \\ & n &=& -1+ 3885b \\ & \mathbf{n} &\mathbf{\equiv}& -1 \mathbf{\pmod{3885}} \qquad (1)\\ \end{array}\\ \hline \end{array}$

$\begin{array}{|l|cll|} \hline \begin{array}{|rcll|} \hline \mathbf{ n } & \mathbf{\equiv} & \mathbf{111 \pmod{121}} \\ \mathbf{ n } & \mathbf{\equiv} & \mathbf{111 \pmod{122}} \\ \hline \end{array} \\ \begin{array}{lrcll} \Rightarrow & n -111 &=& 121x \\ & n -111 &=& 122y \\\\ \Rightarrow & n &=& 111+121x \\ & n &=& 111+122y \\\\ & n=111x+121y&=& 111+122y \\ & 111x+121y&=& 111+122y \\\\ & \mathbf{121x-122y}&\mathbf{=}& \mathbf{0} \qquad x,y \in Z \\ &\rightarrow \quad x &=& 122a^{^2)} \qquad a \in Z \\\\ & n &=& 111+121x \\ & n &=& 111+121(122a) \\ & n &=& 111+121\cdot 122a \\ & n &=& 111+14762a \\ & \mathbf{n} &\mathbf{\equiv}& 111 \mathbf{\pmod{14762}} \qquad (2)\\ \end{array} \\ \hline \end{array}$

After reducing, we have two formulas:

$\begin{array}{|rcll|} \hline \mathbf{n} &\mathbf{\equiv}& -1 \mathbf{\pmod{3885}} \qquad &(1)\\ \mathbf{n} &\mathbf{\equiv}& 111 \mathbf{\pmod{14762}} \qquad &(2)\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{n} &\mathbf{\equiv}& -1 \mathbf{\pmod{3885}} \\ \mathbf{n} &\mathbf{\equiv}& 111 \mathbf{\pmod{14762}} \\ \hline \end{array} \\ \begin{array}{lrcll} \Rightarrow & n +1 &=& 3885x \\ & n -111 &=& 14762y \\\\ \Rightarrow & n &=& -1+3885x \\ & n &=& 111+14762y \\\\ & n=-1+3885x&=& 111+14762y \\ & -1+3885x&=& 111+14762y \\ & \mathbf{3885x-14762y}& \mathbf{=}& \mathbf{112} \qquad x,y \in Z \\ &\rightarrow \quad x &=&3378+14762g^{^3)} \qquad g \in Z \\\\ & n &=& -1+3885x \\ & n &=& -1+3885\cdot (3378+14762g) \\ & n &=& -1+3885\cdot 3378+ 3885\cdot 14762g \\ & \mathbf{n} &\mathbf{=}& \mathbf{13123529 + 57350370g \qquad g \in Z } \\ \end{array}$

The smallest positive integer is  $\mathbf{13\ 123\ 529}$

Proof:

$\begin{array}{|rcll|} \hline \mathbf{13\ 123\ 529} &\text{mod } 105 =& 104\\ \mathbf{13\ 123\ 529} &\text{mod } 111 =& 110\\ \mathbf{13\ 123\ 529} &\text{mod } 121 =& 111 \\ \mathbf{13\ 123\ 529} &\text{mod } 122 =& 111 \\ \hline \end{array}$

$\begin{array}{lcll} \hline ^1) \\ \text{Solve of the diophantine equation $35x - 37y = 2$ } \\ \text{The variable with the smallest coefficient is $x$. The equation is transformed after $x$: }\\ \begin{array}{rcll} 35x &=& 2 + 37y \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 2 + 37y } {35}} \\ &=& \dfrac{ 2 +35y+2y } {35} \\ &=& \dfrac{ 35y + 2 +2y } {35} \\ &=& \dfrac{35y}{35} + \dfrac{ 2 +2y } {35} \\ x &=& y + \dfrac{ 2 +2y} {35} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{ 2 + 2y } {35} \\ & 35a &=& 2 + 2y \\ \end{array} \\ \text{The variable with the smallest coefficient is $y$. The equation is transformed after $y$: }\\ \begin{array}{rcll} 2y &=& -2 + 35a \\ \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -2 + 35a } {2}} \\ &=& \dfrac{ -2 + 34a+a } {2} \\ &=& \dfrac{ 34a - 2 +a } {2} \\ &=& -\dfrac{-2}{2} +\dfrac{34a}{2} + \dfrac{ a } {2} \\ y &=& -1+17a+ \dfrac{ a } {2} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &b &=& \dfrac{a} {2} \\ & 2b &=& a \\ \end{array} \\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{a} &\mathbf{=}& \mathbf{ 2b } \\ \end{array} \end{array}$

$\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -2 + 35a } {2}} \quad & | \quad \mathbf{a = 2b }\\ & = & \dfrac{-2 + 35\cdot (2b) } {2} \\ \mathbf{y} & \mathbf{=} & \mathbf{-1+35b} \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 2 + 37y } {35}} \quad & | \quad \mathbf{y = -1+35b }\\ & = & \dfrac{ 2 + 37\cdot (-1+35b ) } {11} \\ \mathbf{x} & \mathbf{=} & \mathbf{-1 + 37b } \\ \end{array}$

$\begin{array}{lcll} \hline ^2) \\ \text{Solve of the diophantine equation $121x - 122y = 0 $ } \\ \text{The variable with the smallest coefficient is $x$. The equation is transformed after $x$: }\\ \begin{array}{rcll} 121x &=& 122y \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 122y } {121}} \\ &=& \dfrac{ 121y+y } {121} \\ &=& \dfrac{121y}{121} + \dfrac{y} {121} \\ x &=& y + \dfrac{ y} {121} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{y} {121} \\ & 121a &=& y \\ \end{array} \\ \text{The variable with the smallest coefficient is $y$. The equation is transformed after $y$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{y} &\mathbf{=}& \mathbf{ 121a } \\ \end{array} \end{array}$

$\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 122y } {121}} \quad & | \quad \mathbf{y=121a }\\ & = & \dfrac{122\cdot (121a) } {121} \\ \mathbf{x} & \mathbf{=} & \mathbf{122a} \\ \end{array}$

$\begin{array}{lcll} \hline ^3) \\ \text{Solve of the diophantine equation $3885x - 14762y = 112 $ } \\ \text{The variable with the smallest coefficient is $x$. The equation is transformed after $x$: }\\ \begin{array}{rcll} 3885x &=& 112 + 14762y \\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 112 + 14762y} {3885}} \\ &=& \dfrac{ 112 + 11655y + 3107y } {3885} \\ &=& \dfrac{11655y + 112 + 3107y }{3885} \\ &=& \dfrac{11655y }{3885} + \dfrac{112 + 3107y} {3885} \\ x &=& 3y + \dfrac{ 112 + 3107y } {3885} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &a &=& \dfrac{112 + 3107y } {3885} \\ & 3885a &=& 112 + 3107y \\ \end{array} \\ \text{The variable with the smallest coefficient is $y$. The equation is transformed after $y$: }\\ \begin{array}{rcll} 3107y &=& -112 + 3885a \\ \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -112 + 3885a} {3107}} \\ &=& \dfrac{ -112 + 3107a + 778a } {3107} \\ &=& \dfrac{3107a - 112 + 778a }{3107} \\ &=& \dfrac{3107a }{3107} + \dfrac{- 112 + 778a} {3107} \\ y &=& 3a + \dfrac{ 112 + 3107a } {3107} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &b &=& \dfrac{-112 + 778a } {3107} \\ & 3107b &=& -112 + 778a \\ \end{array} \\ \text{The variable with the smallest coefficient is $a$. The equation is transformed after $a$: }\\ \begin{array}{rcll} 778a &=& 112 + 3107b \\ \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ 112 + 3107b} {778}} \\ &=& \dfrac{ 112 + 2334b + 773b } {778} \\ &=& \dfrac{2334b +112 + 773b }{778} \\ &=& \dfrac{2334b }{3107} + \dfrac{112 + 773b} {778} \\ a &=& 3b + \dfrac{ 112 + 773b} {778} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &c &=& \dfrac{112 + 773b } {778} \\ & 778c &=& 112 + 773b \\ \end{array} \\ \text{The variable with the smallest coefficient is $b$. The equation is transformed after $b$: }\\ \begin{array}{rcll} 773b &=& -112 + 778c\\ \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ -112 + 778c} {773}} \\ &=& \dfrac{ -112 + 773c + 5c } {773} \\ &=& \dfrac{773c -112 + 5c }{773} \\ &=& \dfrac{773c }{773} + \dfrac{-112 + 5c } {773} \\ b &=& c + \dfrac{ -112 + 5c} {773} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &d &=& \dfrac{-112 + 5c } {773} \\ & 773d &=& -112 + 5c \\ \end{array} \\ \text{The variable with the smallest coefficient is $c$. The equation is transformed after $c$: }\\ \begin{array}{rcll} 5c &=& 112 + 773d\\ \mathbf{c} &\mathbf{=}& \mathbf{\dfrac{ 112 + 773d } {5}} \\ &=& \dfrac{ 110+2 + 770d + 3d } {5} \\ &=& \dfrac{110+770d+2 + 5d }{5} \\ &=& \dfrac{110 }{5}+\dfrac{770d }{5} + \dfrac{2 + 3d } {5} \\ c &=& 22 + 154d + \dfrac{2 + 3d } {5} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &e &=& \dfrac{ 2 + 3d } {5} \\ & 5e &=& 2 + 3d \\ \end{array} \\ \text{The variable with the smallest coefficient is $d$. The equation is transformed after $d$: }\\ \begin{array}{rcll} 3d &=& -2 + 5e\\ \mathbf{d} &\mathbf{=}& \mathbf{\dfrac{ -2 + 5e } {3}} \\ &=& \dfrac{ -2 + 3e + 2e } {3} \\ &=& \dfrac{3e -2 + 2e}{3} \\ &=& \dfrac{3e }{3} + \dfrac{-2 + 2e} {3} \\ d &=& e + \dfrac{-2 + 2e } {3} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &f &=& \dfrac{ -2 + 2e } {3} \\ & 3f &=& -2 + 2e \\ \end{array} \\ \text{The variable with the smallest coefficient is $e$. The equation is transformed after $e$: }\\ \begin{array}{rcll} 2e &=& 2 + 3f\\ \mathbf{e} &\mathbf{=}& \mathbf{\dfrac{2 + 3f } {2}} \\ &=& \dfrac{ 2+2f+f } {2} \\ &=& \dfrac{2 }{2} + \dfrac{2f }{2} + \dfrac{f} {2} \\ e &=& 1 + f + \dfrac{f} {2} \\ \end{array}\\ \begin{array}{lrcll} \text{we set:} &g &=& \dfrac{ f} {2} \\ & 2g &=& f \\ \end{array} \\ \text{The variable with the smallest coefficient is $f$. The equation is transformed after $f$: }\\ \begin{array}{lrcll} \text{no fraction there:} & \mathbf{f} &\mathbf{=}& \mathbf{ 2g } \\ \end{array} \end{array}$

$\text{Elemination of the unknowns:}\\ \begin{array}{rcll} \mathbf{e} &\mathbf{=}& \mathbf{\dfrac{ 2 + 3f } {2}} \quad & | \quad \mathbf{f=2g }\\ & = & \dfrac{2 + 3\cdot 2g } {2} \\ \mathbf{e} & \mathbf{=} & \mathbf{1 + 3g } \\\\ \mathbf{d} &\mathbf{=}& \mathbf{\dfrac{ -2 + 5e } {3}} \quad & | \quad \mathbf{e=1 + 3g }\\ & = & \dfrac{-2 + 5\cdot (1 + 3g) } {2} \\ \mathbf{d} & \mathbf{=} & \mathbf{1 + 5g } \\\\ \mathbf{c} &\mathbf{=}& \mathbf{\dfrac{ 112 + 773d } {5}} \quad & | \quad \mathbf{d=1 + 5g }\\ & = & \dfrac{112 + 773\cdot (1 + 5g) } {2} \\ \mathbf{c} & \mathbf{=} & \mathbf{177 + 773g } \\\\ \mathbf{b} &\mathbf{=}& \mathbf{\dfrac{ -112 + 778c } {773}} \quad & | \quad \mathbf{c=177 + 773g }\\ & = & \dfrac{-112 + 778\cdot (177 + 773g) } {773} \\ \mathbf{b} & \mathbf{=} & \mathbf{ 178 + 778g } \\\\ \mathbf{a} &\mathbf{=}& \mathbf{\dfrac{ 112 + 3107b } {778}} \quad & | \quad \mathbf{b=178 + 778g }\\ & = & \dfrac{112 + 3107\cdot (178 + 778g) } {778} \\ \mathbf{a} & \mathbf{=} & \mathbf{ 711 + 3107g } \\\\ \mathbf{y} &\mathbf{=}& \mathbf{\dfrac{ -112 + 3885a } {3107}} \quad & | \quad \mathbf{a=711 + 3107g }\\ & = & \dfrac{-112 + 3885\cdot (711 + 3107g) } {3107} \\ \mathbf{y} & \mathbf{=} & \mathbf{ 889 + 3885g } \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{ 112 + 14762y } {3885}} \quad & | \quad \mathbf{y=889 + 3885g }\\ & = & \dfrac{112 + 14762\cdot (889 + 3885g) } {3885} \\ \mathbf{x} & \mathbf{=} & \mathbf{ 3378 + 14762g } \\ \end{array}$

 আগের মতই বিকেলের ম্যাচগুলো বাংলাদেশ সময় সাড়ে ৪টায়। আর সন্ধ্যার ম্যাচগুলো রাত সাড়ে ৮টায় শুরু হবে।
এবার লিগ পর্যায়ের মোট ১৩টি ডাবল হেডার ম্যাচ খেলা হবে শনি ও রোববারে আর ৪৪টি ম্যাচ হবে সন্ধ্যায়। এছাড়া ১২টি ম্যাচ হবে বিকেলে।

এদিকে গভর্নিং কাউন্সিলের সভা শেষে আরও জানানো হয়েছে, চলতি আসরে কিংস ইলেভেন পাঞ্জাব তাদের চারটি হোম ম্যাচ খেলবে মোহালিতে, বাকি তিনটি হবে ইন্দোরে। ফলে নবম ভেন্যু হিসেবে অভিষেক হচ্ছে ইন্দোরের।

www.youtube.com/channel/UCbu-Lu9mfKXHpDzLrneJmUg

$7,560 - $100 = $7,460 Balance of the loan owed by Scott.

$7,460 x 7.2% x 18/12 =

$7,460 x 0.072 x 1.5 = $805.68 - simple interest on the loan for 18 months, or 18/12 = 1.5 years.

$7,460 + $100 + $805.68 = $8,365.68 - What Scott paid for the ring.

3 2/7 > 0.7/0.2        You got this one wrong!

0.7/0.2 = 3 1/2         Therefore:

3 2/7 < 3 1/2

4 + 7*8 - (4 -5) =       You got this one right!

4 + 56 - (-1) =

4 + 56 + 1 =61

What effect does replacing x with  x−4 have on the graph for the function f(x)?

f(x)=|x−6|+2

The graph is shifted 4 units left.

The graph is shifted 4 units right.

The graph is shifted 4 units up.

The graph is shifted 4 units down.

warm and wary

Here is a graph of a standard bell curve:  (ya just have to memorize the divisions)

6.5 m would be at the peak

Each line represents an oarfish 1.5 m shorter

The 16% total line is over to the left.....One line from the mean   6.5 - 1 line = 6.5-1.5 = 5 m

....so 16% are shorter than 5m