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Thank you!!!! Very well explained and I now know how to do it well!

Yes, I got   4/3 too.

See if you can work it out from my pic.

Let TU and VW be chords of a circle, which intersect at S, as shown. If ST = 3, TU = 15, and VW = 3, then find SW.

 ^{No problem!}

There are three squares inside the triangle .

Find the area of the third triangle with steps , please

\(\begin{array}{|rcl|} \hline \tan(\varphi) &=& \dfrac{4}{p_1} \qquad \tan(\varphi) = \dfrac{q_1}{4} \\ \hline p_1+q_1 &=& 4\left(\dfrac{1}{\tan(\varphi)}+\tan(\varphi) \right) \\ \hline \end{array} \begin{array}{|rclcrcl|} \hline \tan(\varphi) &=& \dfrac{3}{p_2} \qquad \tan(\varphi) = \dfrac{q_2}{3} \\ \hline p_2+q_2 &=& 3\left(\dfrac{1}{\tan(\varphi)}+\tan(\varphi) \right) \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \sin(\varphi)&=& \dfrac{x}{p_1+q_1+4} \\ \sin(\varphi)&=& \dfrac{x}{4\left(\dfrac{1}{\tan(\varphi)}+\tan(\varphi) \right)+4} \\ 4\sin(\varphi)&=& \dfrac{x}{ \dfrac{1}{\tan(\varphi)}+\tan(\varphi) +1} \\ \dfrac{1}{\tan(\varphi)}+\tan(\varphi)+1 &=& \dfrac{x}{4\sin(\varphi)}\\ \hline \end{array} \begin{array}{|rcll|} \hline \cos(\varphi)&=& \dfrac{x}{p_2+q_2+3} \\ \cos(\varphi)&=& \dfrac{x}{3\left(\dfrac{1}{\tan(\varphi)}+\tan(\varphi) \right)+3} \\ 3\cos(\varphi)&=& \dfrac{x}{\dfrac{1}{\tan(\varphi)}+\tan(\varphi) +1} \\ \dfrac{1}{\tan(\varphi)}+\tan(\varphi) +1 &=& \dfrac{x}{3\cos(\varphi)}\\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \dfrac{x}{4\sin(\varphi)} &=& \dfrac{x}{3\cos(\varphi)}\\ \mathbf{\tan(\varphi)} &=& \mathbf{\dfrac{3}{4}} \\ && \Rightarrow \cos(\varphi) = \dfrac{4}{\sqrt{3^2+4^2}} \\ && \Rightarrow \mathbf{\cos(\varphi) = \dfrac{4}{5} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x &=& 3\cos(\varphi) \left(\dfrac{1}{\tan(\varphi)}+\tan(\varphi) +1\right) \\\\ x &=& \dfrac{3*4}{5} \left(\dfrac{1}{ \dfrac{3}{4} }+\dfrac{3}{4} +1\right) \\\\ x &=& \dfrac{3*4}{5} \left(\dfrac{4}{3}+\dfrac{3}{4} +1\right) \\\\ x &=& \dfrac{4}{5} \left(4+\dfrac{9}{4} +3\right) \\\\ x &=& \dfrac{4}{5} \left(7+\dfrac{9}{4} \right) \\\\ x &=& \dfrac{4}{5} * \dfrac{37}{4} \\\\ x &=& \dfrac{37}{5} \\\\ \mathbf{x} &=& \mathbf{7.4} \\ \mathbf{x^2} &=& \mathbf{54.76\ \text{cm}^2} \\ \hline \end{array}\)

The area of the third triangle is \(\mathbf{54.76\ \text{cm}^2}\)

I have 8 pieces of strawberry candy (all identical) and 7 pieces of rhubarb candy (all identical). Find the number of ways I can distribute this candy to 4 children.

Are the 4 children all identical? I will assume not.

I will also assume that 1 or more children could get nothing.

Say you put the strawberries in a line.  You put one child at the front of the line.   Now you insert the other 3 children somewhere in the line.

Each child gets the strawberries that are between them and the next child.

The first child is set

So we need to know how many variations there are with 3 children and 8 strawberries.  That is 11C3 = 11C8 = 165

(this is called stars and bars method)

For the rhubarbs it is    10C3 =  10C7 = 120

Put them together and you get 120*165 = 19800  ways.

Why didn't you present your post properly before you published it?

If our kind guest had not already answered I would have removed the question from the forum.

The area of a triangle can be obtained from Heron's formula:

\(Area = \sqrt{s(s-a)(s-b)(s-c)}\)

where a, b and c are the lengths of the sides and s = (a+b+c)/2

In this case a = b = c = 12/3 = 4 and s = 12/2 = 6 so

\(Area = \sqrt{6\times 2^3}=\sqrt{48}=4\sqrt3\)

I'll leave you to turn this into a decimal to the nearest tenth.

angle x = 10º

angle y = 10º

angle z = 160º

For the graph of a certain quadratic y=ax^2+bx+c, the vertex of the parabola is (3,7) and one of the x-intercepts is (-2,0) . What is the x-coordinate of the other x-intercept?

Find the quadratic f(x) = ax^2 + bx + c such that f(1) = 3, f(-1) = 7, and f(5) = 26.

Simplify 

\(0^2 \dbinom{n}{0} + 1^2 \dbinom{n}{1} + 2^2 \dbinom{n}{2} + 2^3 \dbinom{n}{2} + \dots + (n-1)^2 \dbinom{n}{n-1}+ n^2 \dbinom{n}{n}\),

where \(n \ge 2\).

\(\begin{array}{|rcll|} \hline \mathbf{S} &=& \mathbf{0^2 \dbinom{n}{0} + 1^2 \dbinom{n}{1} + 2^2 \dbinom{n}{2} + 2^3 \dbinom{n}{2} + \dots + (n-1)^2 \dbinom{n}{n-1}+ n^2 \dbinom{n}{n} } \\\\ \mathbf{S} &=& \mathbf{\sum \limits_{k=0}^{n} k^2\dbinom{n}{k} } \\\\ S &=& \sum \limits_{k=0}^{n} k*k*\dbinom{n}{k} \quad | \quad \mathbf{k = \dbinom{k}{1}} \\\\ S &=& \sum \limits_{k=0}^{n} k\dbinom{k}{1}\dbinom{n}{k} \quad | \quad \color{red}\dbinom{k}{1}\dbinom{n}{k}= \dbinom{n}{1}\dbinom{n-1}{k-1} \\\\ S &=& \sum \limits_{k=1}^{n} k\dbinom{n}{1}\dbinom{n-1}{k-1} \quad | \quad \mathbf{\dbinom{n}{1}=n } \\\\ S &=& \sum \limits_{k=1}^{n} kn\dbinom{n-1}{k-1} \\\\ \mathbf{S} &=& \mathbf{ n\sum \limits_{k=1}^{n} k\dbinom{n-1}{k-1} }\\ \hline \end{array}\)

\(\mathbf{ \sum \limits_{k=1}^{n} k\dbinom{n-1}{k-1} = \ ? }\)

\(\begin{array}{|rcll|} \hline \mathbf{\left( 1+x \right)^{n-1}} &=& \mathbf{\dbinom{n-1}{0}x^0 +\dbinom{n-1}{1}x^1 + \dbinom{n-1}{2}x^2 + \dots + \dbinom{n-1}{n-1}x^{n-1}} \quad | \quad *x \\\\ x\left( 1+x \right)^{n-1} &=& \dbinom{n-1}{0}x^1 +\dbinom{n-1}{1}x^2 + \dbinom{n-1}{2}x^3 + \dots + \dbinom{n-1}{n-1}x^{n} \\\\ x\left( 1+x \right)^{n-1} &=& \sum \limits_{k=0}^{n-1} \dbinom{n-1}{k}x^{k+1} \\\\ \dfrac{d}{dx}x\left( 1+x \right)^{n-1} &=& \sum \limits_{k=0}^{n-1} (k+1) \dbinom{n-1}{k}x^{k} \\\\ x(n-1) \left(1+x \right)^{n-2} + \left( 1+x \right)^{n-1} &=& \sum \limits_{k=0}^{n-1} (k+1) \dbinom{n-1}{k}x^{k} \quad | \quad \mathbf{x=1} \\\\ (n-1) \left(1+1 \right)^{n-2} + \left( 1+1 \right)^{n-1} &=& \sum \limits_{k=0}^{n-1} (k+1) \dbinom{n-1}{k}1^{k} \\\\ (n-1) \left(2 \right)^{n-2} + \left( 2 \right)^{n-1} &=& \sum \limits_{k=0}^{n-1} (k+1) \dbinom{n-1}{k} \\\\ (n-1) 2^{n-2} + 2^{n-1} &=& \mathbf{\sum \limits_{k=1}^{n} k \dbinom{n-1}{k-1}} \\ \hline \end{array}\)

\(\mathbf{S= \ ?} \)

\(\begin{array}{|rcll|} \hline \mathbf{S} &=& \mathbf{ n\sum \limits_{k=1}^{n} k\dbinom{n-1}{k-1} } \quad | \quad \mathbf{\sum \limits_{k=1}^{n} k \dbinom{n-1}{k-1} = (n-1) 2^{n-2} + 2^{n-1} }\\\\ S &=& n\left( (n-1) 2^{n-2} + 2^{n-1} \right) \\\\ S &=& n\left( (n-1) 2^{n-2} + 2^{n-1}*\dfrac{2}{2} \right) \\\\ S &=& n\left( (n-1) 2^{n-2} + 2^{n-2}*2 \right) \\\\ S &=& n2^{n-2}\left( n-1 + 2 \right) \\\\ S &=& n2^{n-2}\left( n+1 \right) \\\\ \mathbf{S} &=& \mathbf{n(n+1)2^{n-2}} \quad | \quad f(n)= n(n+1),\ g(n) = n-2,\ n \ge 2\\ \hline \end{array}\)

One question per post.

your thread has been closed.

Suppose that the height (in centimeters) of a candle is a linear function of the amount of time (in hours) it has been burning. After 9 hours of burning, a candle has a height of 25.4 centimeters. After 23 hours of burning, its height is 19.8 centimeters. What is the height of the candle after 22 hours?

The distance is  1 1/3  

Let O be a center of a semicircle and T the tangent point on XY

∠XYO = arccos( YO / XY )

Radius  TO = sin( ∠XYO ) * YO 

Area = 1/2 (TO² *pi )  

Randomly choose two numbers between 0 and 1.
What is the probability that their mean is no greater than 2/3?

\(\text{Let number one $= x$ } \\ \text{Let number two $= y$ } \\ \text{Let the mean $= \dfrac{x+y}{2}$ } \)

Their mean is no greater than \(\dfrac{2}{3}\):
\(\dfrac{x+y}{2} \le \dfrac{2}{3}\)

\(\begin{array}{|rcll|} \hline \dfrac{x+y}{2} &=& \dfrac{2}{3} \\ x+y &=& \dfrac{4}{3} \quad | \quad y=1 \\ x+1 &=& \dfrac{4}{3} \\ x &=& \dfrac{4}{3}-1 \\ \mathbf{x} &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{x+y}{2} &=& \dfrac{2}{3} \\ x+y &=& \dfrac{4}{3} \quad | \quad x=1 \\ 1+y &=& \dfrac{4}{3} \\ y &=& \dfrac{4}{3}-1 \\ \mathbf{y} &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{The probability} &=& \dfrac{A_{\color{red}red}}{A} \\\\ && A_{\color{red}red} = \dfrac{\left(1+\frac13\right)}{2}*\dfrac23+\dfrac13*1 \\\\ && \mathbf{A_{\color{red}red} = \dfrac{7}{9}} \\\\ && A = 1*1 \\\\ && \mathbf{A = 1} \\\\ \text{The probability} &=& \dfrac{\frac{7}{9}}{1} \\\\ \mathbf{\text{The probability}} &=& \mathbf{\dfrac{7}{9}} \\ \hline \end{array}\)

The sides of triangle XYZ are XY = XZ = 25 and YZ = 40.

A semicircle is inscribed in triangle XYZ so that its diameter lies on \(\overline{YZ}\), and is tangent to the other two sides.

Find the area of the semicircle.

When I realized that the 2 triangles between purple squares have sides 3-4-5, it was a piece of cake to solve it.

The yellow square area is  54.76 cm² 

Hint:

I have not looked properly but

You probably need this.  

As for you being grateful, I will believe it when I start seeing it.

It doesn't matter anyway because this question does not make sense.

You probably need information from the message board in order to understand what is even being asked.

1)  Extend line segment QS to point T on a circle.

QT = PQ² / RQ              RT = QT - RQ         

M is a midpoint of line segment RT. Now we have a right triangle OMS.

MO = sqrt( OS² - MS² )

Radius  OR = sqrt( MR² + MO² )

Please see attached. There is a full and detailed explanation there.

oh.... sorry about that....

Can you pls just explain me the steps??😁😁😁