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Area of square (or rectangle) = Length x Width  units^2    (for a SQUARE   L & W are equal)

Area of a triangle = 1/2 Base x height   units^2

Area of a rectangle = L  x  W

                                = 7.3 x 4.1   m^2      OK?   (I do not know what the 9. is in your question...typo?) 

Thank you!

The points $B(1, 1)$, $I(2, 4)$ and $G(5, 1)$ are plotted in the standard rectangular coordinate system to form triangle $BIG$.

Triangle $BIG$ is translated five units to the left and two units upward to triangle $B'I'G'$,

in such a way that $B'$ is the image of $B$, $I'$ is the image of $I$, and $G'$ is the image of $G$.

What is the midpoint of segment $B'G'$?

Express your answer as an ordered pair.

\(\begin{array}{|l|lrr|} \hline & \text{translation} \\ & x' =x-5\\ & y' = y +2 & \\ \hline \text{B} =(1,1) & \text{B'} =(1-5,1+2) = (-4,3) \\ \text{I} = (2,4) & \text{I'} = (2-5,4+2) = (-3,6) \\ \text{G} = (5,1) & \text{G'} = (5-5,1+2) = (0,3) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{midpoint of } B'G' &=& \dfrac{B'+G'}{2} \\\\ &=& \dfrac{\dbinom{-4}{3}+\dbinom{0}{3}}{2} \\\\ &=& \dfrac{\dbinom{-4+0}{3+3}}{2} \\\\ &=& \dfrac{\dbinom{-4}{6}}{2} \\\\ &\mathbf{=}& \mathbf{ \dbinom{-2}{3} } \\ \hline \end{array}\)

Your Psychosis seems to have flared up dramatically in the last couple of weeks or so. My advice to you: see Jordan Peterson's talk on "Crazy Women!".

Yes, I saw it. I’m not going to reproduce it because I don’t have enough Dramamine.

However, here’s a paraphrase of the FAKE MATH highlights.

You wrote: sqrt(225)  =  25, so the radius of the lower base is 25.

(The last time I checked 15 was the sqrt of 225)

You wrote, the height of the big triangle is 24

(It’s actually 24 + h)

You used this to “solve” it

h/24 = 5/25

 h  =  4.8

The only thing interesting about your post is the coincidental "correct" result.

You are user TPM. This is a DNA match for the crap he'd post!

You used FAKE MATH, Blarney, and BS, and you just dumb-lucked into Hectictar’s correct answer, in time to replace your BS with more BS.

GA

This is hilarious!

The reason I thought you were Mr. BB, is your presentation of this formula/equation

[3 + 3 - 1] nCr 3 =5C3 = 5! / [5 - 3]!. 3! = 10

in irregular mathematical syntax. I usually refer to this style as SLOP, because that is what it is. (It looks like the factorial of 3 = 10)

4 is the greatest common factor:

.

What is the product of all the coordinates of all the points of intersection of the two circles defined by x^2-2x+y^2-10y+25=0 and x^2-8x+y^2-10y+37=0?

The radius of the first cone is 5 meters,

so the diameter of the first cone is 10 meters.

The diameter of the second cone is 15 meters.

\(\frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\Big(\frac{\text{diameter of second cone}}{\text{diameter of first cone}}\Big)^3\\~\\ \frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\Big(\frac{15}{10}\Big)^3\\~\\ \frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\Big(\frac{3}{2}\Big)^3\\~\\ \frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\Big(\frac{3}{2}\Big)\Big(\frac{3}{2}\Big)\Big(\frac{3}{2}\Big)\\~\\ \frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\frac{27}{8}\\~\\ {\small\text{volume of second cone}}\,=\,\frac{27}{8}\cdot{\small\text{volume of first cone}}\)

Assuming that both haystacks have the same density, and each piece of hay weighs the same....

\({\small\text{weight of second cone}}\,=\,\frac{27}{8}\cdot{\small\text{weight of first cone}}\\~\\ {\small\text{weight of second cone}}\,=\,\frac{27}{8}\cdot12 \text{ tons}\\~\\ {\small\text{weight of second cone}}\,=\,40.5 \text{ tons}\)

I don't know the correct way to solve this, but first I guessed that x is 8 point something,

which makes  floor(x) = 8

And   70 / 8  =  8.75    , so this works:       \( ​​​​​​\lfloor 8.75 \rfloor \cdot 8.75 = 70 \)           

sin 60 = AB/4         (sin = opposite/hypotenuse)

sqrt3 / 2 = AB/4

4 sqrt3 / 2 = AB

AB= 2 sqrt(3)

Are the 7 players taking biology ONLY...or are the 2 players taking 'BOTH' included in this 7 count?

Unclear as how to proceed.....

Need a diagram for this one.....or some measurements are missing.  Sorry.....

  I have no idea who Mr BB is, WildChild.... I just deleted an incorrect answer I had posted as I was responding with a different (at least my second 'guess' was correct) answer to another poster.

G'Day!  

You shouldn’t delete your posts, Mr. BB.

According to experts, there should be:

[3 + 3 - 1] nCr 3 =5C3 = 5! / [5 - 3]!. 3! = 10 ways of putting the lamps on the tables.

You should just not make them in the first place!

You should also update your “expert” consultants list.  If you put my name at the top of the list,  then you will always have the correct answers.

The difference of twice the number t and 30 equals 2t - 30

2t-30

If they sit around the table ....can they not be seated   

3    2                          2      3

   1              and             1            I don't think  these can be rotated to produce the other...so there would be   288 * 2   ways to be seated around the table.    

A partition set for a number does not normally have empty sets, but a table can have zero lamps, so by implication there is a zero. 

There are three sets for this question, and the question allows for empty sets.

GA

OK. I’m an expert.

The formula says: P[n, k] =P[10, 3] =P[7,1] + P[7, 2] + P[7, 3] = 8 different ways as follows:

The formula is correct, but it is incomplete for this question.

P stands for "Partitions", which means "How many ways can you make up a 10 out of 3 numbers from 1 to 10". So, I calculated it as being 8 ways: 1+1+8, 1+2+7, 1+3+6, 1+4+5, 2+2+6, 2+3+5, 2+4+4, 3+4+3. And that gives 8 ways of making a 10 out of 3 numbers from 1 to 10!!!!!!.

This is the correct interpretation for the incomplete formula.   

Note: 1+1+8, 1+8+1, 8+1+1 are considered all the same or just 1 combination.....etc.

There are no combinations here, only partitions. 

No banana for you. 

GA

The general formula:

\(\large \sum\limits_{i=1}^{T}P(L, i) \tiny \leftarrow \text{P is the partition(s) of i in sets of L; T = # of tables, L = # of lamps, and i is the index of T}\\\\ \large \sum\limits_{i=1}^{10}P(3, i) = 14 \tiny \leftarrow \text{With specified parameters. }\\ \)

Descriptive list:

10 on one table & none on the others

9 on one table & 1 on another & none on the other

8 on one table & 2 on another & none on the other

7 on one table & 3 on another & none on the other

6 on one table & 4 on another & none on the other

5 on one table & 5 on another & none on the other

8 on one table & 1 on another & 1 on the other

7 on one table & 2 on another & 1 on the other

6 on one table & 3 on another & 1 on the other

6 on one table & 2 on another & 2 on the other

5 on one table & 4 on another & 1 on the other

5 on one table & 3 on another & 2 on the other

4 on one table & 4 on another & 2 on the other

4 on one table & 3 on another & 3 on the other

GA

The interior angle of a regular polygon can be determined with the following formula:

\(\frac{180(n-2)}{n}\) where n represents the number of sides the polygon has

The measure of each exterior angle of a regular polygon can be determined using the following formula:

\(\frac{360}{n}\) where n represents the number of sides the polygon has

According to the problem, "the measure of each interior angle is four times the measure of each exterior angle," so we can create an equation for this like the following:

This question requires some observation combined with knowledge of properties of parallelograms and triangle similarity.

Parallelograms, by definition, are quadrilaterals with opposite sides parallel. In this example, \(\overline{AD}\parallel \overline {FC}\) because of the previous statement. Because \(\overline{DF}\) intersects multiple segments, namely \(\overline{AD}\text{ and } \overline {FC}\), \(\overline{DF}\) is classified as a transversal. \(\angle ADF\text{ and }\angle F\) are a pair of angles known as alternate interior angles because they exist on opposite sides of the transversal while remaining on the "inner" side of the intersecting lines. Because \(\overline{AD}\parallel \overline {FC}\), the Alternate Interior Angles Theorem states that \(\angle ADF \cong \angle F\). \(\overline{AB}\) is another transversal bounded by parallel lines. Therefore, \(\angle A\cong \angle FBE\). Because there are two pairs of angles that are congruent in corresponding triangles, namely \(\angle ADF \cong \angle F\text{ and }\angle A\cong \angle FBE\), the Angle-Angle Similarity Theorem states that the triangles are similar. In this case, \(\angle AED\sim \triangle BEF\)

Even after all the logic above, I still do not any side lengths yet. There exists a constant of proportionality between similar triangles. Since the area of \(\triangle BEF\) requires an enlargement by a factor of 9 to obtain the area of \(\angle AED\), the constant of proportionality would be the square root of this. \(\sqrt{9}=3\). Therefore, each side length on the larger triangle is enlarged by a factor of 3. Because \(\overline{BF}\) and \(\overline{AD}\) are corresponding sides according to the triangle similarity statement and \(\overline{AD}\) is enlarged by 3, \(AD=3BF\). 

One property of parallelograms is that opposite side lengths are also congruent. Because \(BC=AD\text{ and }AD=3BF \), \(BC=3FC\). 

Using the principle of the Segment Addition Postulate \(FC=BF+BC=BF+3BF=4BF\). 

Since the constant of proportionality for \(\triangle BEF\text{ and }\triangle CDF\) is 4, the area of \(\triangle CDF\) is 16. 

16-1=15, which is the area of the quadrilateral \(BCDE\). 

\(A_{\text{ABCD}}=9+15=24\text{units}^2\)

The diameter of the circle would be 2(6)=12; thus the longest side of the triangle, the base, would be also 12cm. Using the formula of bh/2 to find the area of triangles, b=12 and h=6, as the height is the radius of the circle. (Would be easier to see on a diagram...)

12x6=72, 72/2=36

The area of the triangle would be 36cm squared