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These angles need to sum to 360 degrees.

120 + 80 + B = 360

B = 160 degrees

legs of a triangle a, b, c

the triangle inequality states

\(|a|+|b| >|c| \\ \text{with strict inequality else we'd just have a line segment, not a triangle.}\)

\(|c| < |39|+|15| = 54 \Rightarrow c < 54 \\ \text{we also must have }\\ |c|+|15| > 39 \Rightarrow c > 24 \text{ thus}\\ 24 < c < 54 \)

\(\text{If I understand what you're saying...}\\ 4\sqrt{abc}=5\\ 4\sqrt{abcd}=3\sqrt{10}\)

\(\dfrac{4\sqrt{abcd}}{4\sqrt{abd}}=\sqrt{d} = \dfrac{3\sqrt{10}}{5}\\ d = \left(\dfrac{3\sqrt{10}}{5}\right)^2 = \dfrac{9\cdot 10}{25} = \dfrac{18}{5}\)

\(\text{assuming a 365 day year}\\ nblink=3\dfrac{blink}{min}\cdot 60\dfrac{min}{hr}\cdot 14\dfrac{hr}{day}\cdot 365 \dfrac{day}{yr} = \\ 3\cdot 60 \cdot 14 \cdot 365 =919800 \dfrac{blink}{yr} \\ \text{so no, not quite a million blinks per year}\)

I think the easiest way to do this is to parameterize the circle via a single variable and solve the minimization problem.

\(\text{let }p \text{ be a point on the circle given by }(x-12)^2 + (y+5)^2 = 9 \\ \text{then p can be described as }\\ p=(3 \cos(\theta)+12,~3\sin(\theta)-5)\)

\(\text{The squared distance of this point to the origin is }\\ d^2 = (3 \cos(\theta)+12)^2 + (3\sin(\theta)-5)^2 =\\ 9 \sin ^2(\theta )-30 \sin (\theta )+9 \cos ^2(\theta )+72 \cos (\theta )+169 = \\ -30 \sin (\theta )+72 \cos (\theta )+178 = \\ 78 \cos(\theta+\arctan(5/12))+178\)

\(d^2 \text{will clearly be minimized when }\\ \theta + \arctan(5/12) = \pi\\ \theta = \pi - \arctan(5/12)\\ \text{at this angle }d^2 = -78 + 178 = 100\\ d=10\)

The key here is to find x such that the determinant of F is zero.

\(det(F) = 2\cdot 3 - (-4)\cdot x = 6+4x \\ \text{solving }det(F)=0 \text{ we get} \\ 6+4x = 0 \\ x = -\dfrac 3 2\)

well let's multiply them

\(\begin{pmatrix}4 &3\\1&1\end{pmatrix} \begin{pmatrix}1 &-3\\-1&4\end{pmatrix} =\begin{pmatrix}1&0\\0&1\end{pmatrix}\)

so yes C and E are multiplicative inverses

Square area = L x L = 52 x 52 = 2704 sq ft

2704 sq ft / (208 sq ft/gal) = 13 gallons

Perimeter: 2l+2w

Area: lw

Perimeter: 2(30)+2(85)=60+170=230 ft

Area: lw=30*85=2550 ft^2

Perhaps you should tell the other guest how you got your answer.

Maybe the person does not know what perimeter or area even mean. 

Here is the formula for the surface are of a cone.

Just plug the values in that you know and solve for h

then us pythagoras and solve for l  (slant height)

Hi Ginger,

5 of the 9 letters of the word MINCEMEAT are selected.

Find the number of possible selections which contain exactly 1 M and 1 E.

I just mean that if 5 letters are selected at random then the prob of drawing MECAT is higher then the prob of drawing INCAT    (becasue there are 2Ms and 2Es to chose from)

So saying there are 10 distinct outcomes is correct but each one  does not have the same probability of occuring as other possible outcomes.

eg

What is the probability of chosing a selection of 5 letters that contain exactly one M and one E

We have determined that there are 10 such unique combinations

How many combinations (not unique) are there altogether.  (think in terms of  M1, M2, E1, E2)

9C5 = 126

Now how many of these have exactly one M and one E

M1 OR M2,  E1 or E2, plus 3 more.

2*2*5C3 = 40

So I think that if 5 letters are selected at random then the prob of having exactly one M and one E is     40/126 = 20/63 

I’ve already answer this question which doesn’t need a answer to because it’s 

perimeter 230 ft

area 2550 ft²

I've played with this a little and Rom's answer is most likely correct but I do not see why.

If I spent hours on proofs I am sure I could show it but I don't see it intuitively at all.

I was surprised when that angle came up as 77 degrees.

There are a total of \(\dbinom{6}{2}=15\) possible angles

But it turns out that many of the angles are congruent.

The 15 angles are made up of 1 + 2 + 3 + 4 + 5 copies of 5 different angles.

There are only 5 different possible angles

LOL

EP, it seems that I’m not the only one who confuses you for the Blarney Banker.

To me, this post is not very consistent with Mr. BB’s style, but we all have our brain-dead moments. These moments are a way-of-life for the forum’s BBs, and rarely is there a discontinuity.

GA

\(\text{This is known as finding the partial fraction expansion}\\ \dfrac{x+7}{x^2-x-2} = \dfrac{x+y}{(x-2)(x+1)} = \dfrac{A}{x-2}+\dfrac{B}{x-1}\)

\(\dfrac{x+7}{(x-2)(x+1)} = \dfrac{A(x+1)+B(x-2)}{(x-2)(x+1)}\\ \forall x : x \neq 2,~-1 \\ x+7 = A(x+1) + B(x-2) \\ x = (A+B)x \Rightarrow \\ 1 = A+B\\ 7 = A -2B \)

\(B = 1-A\\ 7 = A - 2(1-A) = 3A-2\\ 3A = 9\\ A=3\\ B = 1-3 = -2\\ \dfrac{x+7}{(x-2)(x+1)}=\dfrac{3}{x-2}-\dfrac{2}{x+1}\)

Hi Melody, how do you justify this statement?

“Of course if you were doing it for a probability question then some of these combinations are more likely to occur than others.”

Why would a question seeking a probability solution change the enumeration, that is make some combinations more likely to occur?

GA

Thanks Guest, now the first one makes total sense.

There are 3 to chose from out of 5 then add in the M and the E.

So, yea that is the correct number of possible arrangements, I'm in total agreement.

For the second question it is saying you can have

ME then chose 3 out of the 5       5C3       or

MEE then chose 2 out of the 5      5C2     or

MME then chose 2 out of the 5      5C2     or

MMEE then chose 1 out of the 5     5C1  

Add those all up and you have the number of ways.

Yes that makes total sense too. I just realized that this second answer is exactly what I did in the first place LOL.

Melody: I think what he means is the answer to the first part is 10. And it is arrived at by 5C3 = 10

And the answer to the second part is 35. And it is arrived by [5C1 + 5C2 +5C2 + 5C3]=5 + 10 + 10 + 10 =35.

it wouldnt work when i put it in desmos should i put i in y= ? form 

I cannot see any scenario where 10*5C3 would be the answer to the first one.

I do not understand what you have written anyway. There is only 2 questions. What are all those combinations in the bracket for?

2x-y has to equal 18

just plug those values in and see which one works.   

Here is the best way for you to check by yourself.

Put your equation into here and you will see for yourself.