All Answers 356000 Answers

We have that

N  =  a₀b^x       

Where N  =  the number of bees after  x days  and  a₀  is the number of bees on day 0

So we have that

36 =  27b³      divide both sides by  27

36/27    = b³

4/3  = b³       take the cube root

∛[4/3]  = b

After 30 days there are

N  = 27*(∛[4/3])³⁰ ≈  479  bees

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Why. Just why. It's just a mean problem and you just don't want to spend two seconds on it. Why. 

The mean of stats in a table is the sum of all the numbers divided by how many numbers there are. That means that the mean of the high temperatures here would be (82+81+88+88+78+72+81+79+84)/9 = 733/9 which is about 81 degrees F, so (D).

-o_o

First, we want to make sure that 3/5 and 1/4 have the same common denominator. Multiply 3/5 by 4/4 and 1/4 by 5/5, we get that 3/5+1/4 = 12/20+5/20 = 17/20. Hence, Kyle used 17/20yds of wire in all.

-o_o

The one before the last choice (Third one, C)

First simplify the equation of the circle to the general formula which is (x-h)^2+(y-k)^2=r^2

so

(x^2+x)+(y^2-6y)-2

Notice both of these we need to complete the square 

so 

(x^2+x+1/4)+(y^2-6y+9)=+2+9+1/4 (Why did we add 9 and 1/4? Because anything we add to one side of an equation must be added to the other side as well as multiply and divide and subtraction) 

so this basically simplifies to

(x+1/2)^2+(y-3)^2=11.25 (Notice we are in the general equation form of the circle's equation which is again, (x-h)^2+(y-k)^2=r^2

Find the center would be finding x and y

so x and y must make the barract=0 in order to be center

x+1/2=0 

x=-1/2

y-3=0

y=3

Radius^2 is 11.25 (Notice it is squared already due to general form) 

So radius is = sqrt(11.25) equal to 3sqrt(5)/2 

The perimeter of a sector of a circle is the sum of the two sides formed by the radii and the length of the included arc. A sector of a particular circle has a perimeter of 28 cm and an area of 49 sq cm. What is the length of the arc of this sector?

No, not quite. There were 27 bees when they were discovered. After 3 days, there were 36 bees. Right?

What was the INCREASE in 3  days? 

36 / 27 = 1.333333..........There was an increase of  33.33% over 3 days. And after 30 days?

30 / 3 = 10 periods of 33.33% INCREASE, because the bees are multiplying "exponentially". That means:

(1.3333333.....)^10 =17.7577 - this is the total growth in the population of the original 27 bees.

So, 27 x 17.7577 =479.458 = ~479 bees after 30 days.

I will give you a hint,

Average (Arithmetic mean) = Sum of terms/Number of terms

Example:

A,B,C,D

The arithmetic mean is (A+B+C+D)/4

-2+x+x^3-x^4

The center is (2,-3) 

radius is equal to sqrt(80) 

why?

Well the general equation of a circle is

(x-h)^2+(y-k)^2=r^2

Where r is the radius 

h,y are real numbers, (x,y determines the center of the circle, notice that x,y in order to be the center of the circle have to make x-h=0 and y-k=0 

so for example in your question,

(x-2)^2+(y+3)^2=80

notice h and k are -2,+3

solve it by the following

x-2=0

x=2 (First coordinate of the center) x coordinate

y+3=0

y=-3 (Second coordinate of the center) y coordinate

As the general equation of the circle says all of this should give you the radius squared.

So to find the radius we take the squareroot of the number simply

in your question

r^2=80

r=sqrt(80) Notice it can't be negative as it is a radius! 

I think 360 is just the number of bees within 30 days, well we have to add the starting bees, so 360+27=387

B)0.5

Mean=sum of data/no. of data

Eg.

Find the mean of A,B = A+B (Their sum)/(2(Their number)) 

So the question already gave you the number of data which is 2105

and the sum of the data which is 1008.08 

just apply the formula

1008.08/2105=0.47.... rounded to the nearest tenth would be 0.5 which is option B 

The volume of this cylinder is 8700m3. Find the length of the radius, r . Give your answer rounded to 3 SF. The height is 30m   

The volume of a cylinder is the area of the base times the height.   V = h • π • r²    ( π is pi, i.e., 3.1416 )

This specific question    8700  =  30 • 3.14216 • r²  

                                      8700  =  94.248 • r²

                                           r²  =  8700 / 94.248  =  92.3097

                                           r  =  sqrt (92.3097)  =  9.6078  =  9.61 meters  

.

I Don't Think thats the case here, Guest.

\(\frac{19\;rad}{1sec}\qquad \frac{2\pi \;rad}{1revolution}\qquad \frac{9\;inchess}{rad} \qquad \frac{?inches}{sec}\\ \\~\\ \frac{9\;inches}{rad}\times \frac{19\;rad}{1sec}=\frac{9*19\;inches}{second}=171inches/sec\)

Thank you, Melody !

Nice work Heureka! 

An aeroplane is observed at the same instant from three stations A,B, C in a horizontal straight line

but not in a vertical plane through the aeroplane.
If \(AB=BC =c\) and the angles of elevation from A,B C are respectively  \(\alpha\), \(\beta\) and \(\gamma\),

prove that the height of the aeroplane is

\(\dfrac{c\sqrt{2}}{\sqrt{\cot^2(\alpha)+\cot^2(\gamma)-2\cot^2(\beta) }}\)

\(\text{The height of the aeroplane $=h $} \\ \text{$F$ is the nadir point of the aeroplane } \\ \text{$r = \overline{AF}$ } \\ \text{$s = \overline{BF}$ } \\ \text{$t = \overline{CF}$ } \\ \text{$\angle CBF = \epsilon$ } \\ \text{$\angle ABF = 180^\circ-\epsilon$ } \)

\(\begin{array}{|rclcrcl|} \hline \cot(\alpha) &=& \dfrac{r}{h} &\text{or}& r &=& h\cot(\alpha) \qquad (1) \\ \cot(\beta) &=& \dfrac{s}{h} &\text{or}& s &=& h\cot(\beta) \qquad (2) \\ \cot(\gamma) &=& \dfrac{t}{h} &\text{or}& t &=& h\cot(\gamma) \qquad (3) \\ \hline \end{array} \)

cos-rule:

\(\begin{array}{|lrcll|} \hline (1) & t^2 &=& s^2+c^2-2sc\cos(\epsilon) \\\\ & r^2 &=& s^2+c^2-2sc\cos(180^\circ-\epsilon) \\ &&& \boxed{ \cos(180^\circ-\epsilon)=- \cos(\epsilon)} \\ (2) & r^2 &=& s^2+c^2+2sc\cos(\epsilon) \\ \hline (2)+(1): & r^2+t^2 &=& s^2+c^2+2sc\cos(\epsilon)+s^2+c^2-2sc\cos(\epsilon)\\ & r^2+t^2 &=& 2s^2+2c^2 \\ & r^2+t^2-2s^2 &=& 2c^2\\ &&& \boxed{ r^2=h^2\cot^2(\alpha)\\ s^2=h^2\cot^2(\beta)\\ t^2=h^2\cot^2(\gamma) }\\ & h^2\cot^2(\alpha)+h^2\cot^2(\gamma)-2h^2\cot^2(\beta) &=& 2c^2 \\ & h^2\left(\cot^2(\alpha)+ \cot^2(\gamma)-2 \cot^2(\beta)\right) &=& 2c^2 \\ & h^2 &=& \dfrac{2c^2 }{ \cot^2(\alpha)+ \cot^2(\gamma)-2 \cot^2(\beta) } \\\\ &\mathbf{ h } &=& \mathbf{ \dfrac{c\sqrt{2} }{ \sqrt{\cot^2(\alpha)+ \cot^2(\gamma)-2 \cot^2(\beta)} } } \\ \hline \end{array}\)

Thanks Alan,  

Yes that helps me understand the question.

It probably doesn't help me answer it but it gives me a starting point.  

Melody, I interpret this as follows:

Imagine A,B and C in a straight line on the ground, with B in the middle. Now picture a vertical plane (not aeroplane!) intersecting that line. That plane does not go through the aeroplane, i.e. the aeroplane is off to one side of the vertical plane.

Does that make it clearer?
 

The difference of the roots of the quadratic equation \(x^2 + bx + c = 0\) is \(|b - 2c|\).
If \(c \neq 0\), then find \(c\) in terms of \(b\).

\(\begin{array}{|lrcll|} \hline & x^2 + \underbrace{ b }_{-(x_1+x_2)}x + \underbrace{ c }_{x_1x_2} &=& 0 \\ \hline & b &=& -(x_1+x_2) \\ & -b &=& x_1+x_2 \\ (1) & \mathbf{x_2} &=& \mathbf{-x_1-b} \\ \hline (2) & \mathbf{c} &=& \mathbf{x_1x_2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline | x_1-x_2| &=& |b - 2c| \quad &| \quad \mathbf{x_2=-x_1-b} \\ | x_1-(-x_1-b)| &=& |b - 2c| \\ | 2x_1+b)| &=& |b - 2c| \quad &| \quad \text{compare}\\\\ 2x_1 &=& -2c \quad &| \quad : 2 \\ \mathbf{x_1} &=& \mathbf{-c} \\\\ \mathbf{x_2} &=& \mathbf{-x_1-b} \quad (1) \quad & | \quad \mathbf{x_1=-c} \\ x_2 &=& \mathbf{-(-c)-b} \\ \mathbf{x_2} &=& \mathbf{c-b} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{c} &=& \mathbf{x_1x_2} \quad (2) \quad & | \quad \mathbf{x_1=-c},\quad \mathbf{x_2} = \mathbf{c-b} \\ c &=& (-c)(c-b)\quad &| \quad : c \qquad c \neq 0\ ! \\ 1 &=& -(c-b) \\ 1 &=& -c+b \\ \mathbf{c} &=& \mathbf{b-1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x_1} &=& \mathbf{-c} \quad & | \quad \mathbf{c=b-1} \\ x_1 &=& -(b-1) \\ \mathbf{x_1} &=& \mathbf{1-b} \\ \\ \mathbf{x_2} &=& \mathbf{c-b} \quad & | \quad \mathbf{c=b-1} \\ x_2 &=& b-1-b \\ \mathbf{x_2} &=& \mathbf{-1} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline | x_1-x_2| &=& |1-b-(-1)| \\ | x_1-x_2| &=& |1-b+1| \\ | x_1-x_2| &=& |2-b| \\ \hline |b-2c| &=& |b-2(b-1)| \\ |b-2c| &=& |b-2b+2)| \\ |b-2c| &=& |2-b|\ \checkmark \\ \hline \end{array}\)

What do you think?