All Answers 358187 Answers

The hour hand  will  move at  a rate  of 

360°  in 12 hrs

30° in  1 hr

30°/ 60  =  .5° in one minute

So  at   5:44.....the hour  hand has  moved   5 (30°) +  44(.5°)  =  [150 + 22]° = 172°   from 12 noon

The minute hand  will move  at a rate of 360°/ 60  = 6° per minute

So at 5:44...it will have moved  [44 * 6]°  = 264°  from the top of the hour

So...the smaller angle formed by the hands  =  [264 - 172]°  =  92°

nevermind got it :<

So I'm not completely doing your homework for you, I'll just get you started.

1. To find out what type of roots a quadratic in the form ax2+bx+c, plug the coeficients into the discriminant: b2-4ac

If this is positive, there are two real solitutions. If it is negative, there are two non real solitutions. If it is 0, then there is one real solitution. You want to find the discriminants of the 5 quadratics to find the discriminant that is 0.

I'll do the first one:

a=4, b=16 and c=-9

162-4(4)(-9)=256-(-144)=400, so the first one is not the answer.

2. I'm going to interperate this one how I think you meant to write it (you didn't put parenthisies around the xy so the y should be in the numerator): \(\frac{x}{y}=\frac{225}{xy}+\frac{y}{x}\) (and this interperation works out nicely)

First I would start by multiplying by xy to remove the varaibles from the denominator:

x²=225+y²

x²-y²=225

(x+y)(x-y)=225

x and y will be integers when (x+y) and (x-y) are factor pairs of 225

The factors of 225 are:

1 and 225

3 and 75

5 and 45

ect.

For every factor pair there will be integers x and y that satisfy the equation

List the remaining factor pairs until you get to 225 and 1 and count them to get the answer.

To find the solitutions that go with the factors 1 and 225 (which you don't need to do), you would solve x+y=1 and x-y=225

thank you so much!! 

\(\text{An $m\times k$ matrix may multiply a $k \times n$ matrix to produce a $m \times n$ matrix}\)

The example you gave is undefined.

Ahh that makes more sense!!  Thank you so much :)) 

I'll give an explanation on why it is.

The red marbles, blue marbles, and green marbles that Tom has can be arranged as RRRRRRBBBBGGG, where R denotes the red marbles, B denotes the blue marbles, and G denotes the green marbles.

Tom has, therefore, 6+4+3=13 marbles, so we would assume that if each ball was its own different color, then there would be 13!(Thirteen Factorial) ways to arrange the balls. However, there are six(6) red balls, four(4) blue balls, and three(3) green balls, so these are duplicates of the three colors(red, blue, and green). Thus, the number of ways to arrange the thirteen balls are (13!) / (6! * 4! * 3!), \(\frac{13!}{6!*4!*3!}\)which is \(4*7*3*5*11*13=\boxed{60060}\) distinct ways that Tom can line the thirteen balls up. 

Adding my assumption to Melody's, I would assume they would need to have asked the question under a registered account... i.e., not posted the question as a Guest. 

.

You should remove 5......if it remains and is chosen, there is no possibility of of a sum of 10

sin (x)  = 2/3      ⇒  cos(x)   =  √[3^2 - 2^2]  / 3  =  √[5]/3

sec (y)  =  5/4  ⇒  cos (y)   =  4/5  ⇒  sin(y)   =  3/5

sin (x + y)   =  sin (x) cos (y)  + cos (x)  sin( y )  =

(2/3)(4/5)  + (√5)/3 (3/5)  =

[ 8   + 3√5 ]                       

_________               

     15                

Suppose with with \(A, B, C \in \mathbb{R}\).

What is \(A\)?
\(\dfrac{1}{x^3-x^2-21x+45}=\dfrac{A}{x+5}+\dfrac{B}{x-3} + \dfrac{C}{(x - 3)^2}\)

\(x^3-x^2-21x+45=(x+5)(x-3)^2\)

\(\begin{array}{|lrcll|} \hline & \mathbf{\dfrac{1}{x^3-x^2-21x+45} } &=&\mathbf{\dfrac{A}{x+5}+\dfrac{B}{x-3} + \dfrac{C}{(x - 3)^2} } \\\\ & \dfrac{1}{(x+5)(x-3)^2} &=&\dfrac{A}{x+5}+\dfrac{B}{x-3} + \dfrac{C}{(x-3)^2} \quad | \quad \times (x+5)(x-3)^2 \\\\ & 1 &=& A(x-3)^2 + B(x+5)(x-3) + C(x+5) \\ \hline x=-5: & 1 &=& A(-5-3)^2 + B(-5+5)(-5-3) + C(-5+5) \\ & 1 &=& A(-8)^2 + B(0)(-8) + C(0) \\ & 1 &=& 64A \\ & \mathbf{A} &=& \mathbf{ \dfrac{1}{64} } \\ \hline \end{array}\)

Also Given,  Total number os students is 170

I will assume that A' means the number of students NOT doing A.  The notation I would use is a A bar.     \(\bar A\)

\(n(\bar A)=170-50-70 = 50\\ n(\bar C)=170-20-70 = 80\\ n(\bar A \cap \bar C)=170-50-70-20=30 \)

Please edit this and write it properly.

I would assume that they would if they requested to be notified in the first place. 

They may not have made this request.

(Plus it is an assumption on my part)

If sin(x) = 2/3 and sec(y) = 5/4 , where x and y lie between 0 and π/2, evaluate sin(x + y).

Thanks guest,

It is really good to see you give the question a go.

Unfortunately, your answer cannot be fully correct because sine is a ratio.  It is not going to be in degrees or radians.

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

x and y are both acute angles.

Consider a right angled triangle with one acue angle equal to x

opp=2

hyp=3

so   adj=sqrt(9-4) = sqrt 5

sinx=2/3        cosx=sqrt5 / 3

Consider a right angled triangle with one acue angle equal to x

cosy=4/5

adj=4

hyp=5

so   opp=sqrt(16) = 4

\(sin(x+y)=sinx cosy+cosxsiny\\ sin(x+y)=\frac{2}{3} \cdot\frac{4}{5}+\frac{\sqrt5}{3}\cdot\frac{4}{5}\\ sin(x+y)=\frac{20-4\sqrt5}{15}\\ sin(x+y)=\frac{4}{15}(5-\sqrt5)\)

4/15(5-sqrt(5)) approx = 0.7370485393333894    This is a ratio, it is not going to be in degrees or radians.

Not a Trap If you are good at mathematics and It wouldn't take you too much effort (or time) also for competitions , it is better to focus on Number theory,series, functions ,geometry etc.. because calculus isn't involved I think.

Learning calculus at age 11 , as I said If it wouldn't distract you, then no problem. Focusing on competitions? Better not for now..

By the way that site you linked is really great :) 

I just went backwards...

:

T=32-3=29 so volume at that time is=24-4=20

T=29-3=26 , v=16

T=23, v=12

T=20, v =8 (answer)

It didn't take time so it could be a good strategy for this specific question.

Using arthimetic sequence(Really isn't needed and takes a lot of time for this specific question but good if the numbers were far from each other)

We don't know the current temperature (Start) 

So let Temperature= T 

To find T , we know that at somepoint the temperature was 32 and volume was 24 

So T=32-3L 

Where did the 3 come from? "For every 3 deg rise in Temp" So going backward from 32 degrees We wouldn't add 3. We subtract  then

so 32-3L

L is a specific number of terms.. For example, If 32 was for example, the 5th term and i want the 2nd term 

therefore 32-3 -3 -3 etc..

Also could be said as: 32-3L

So Now we know that temperature is equal to , T=32-3L

Same thing goes for the volume.

V=The point which was given with T (t=32 degrees, v=24 cubic cen v) -4L

So 

V=24-4L

where did the 4 come from?

for every 3 degrees, the volume expands by 4.

Going backwards, for decreasing 3 degrees, the volume shrinks(Decreases) by 4

So 24-4L 

L no. of terms backwards as well. 

We now know that:

T=32-3L

V=24-4L

Question:

"What was the volume of the gas in cubic cent. when the temperature was 20 degrees?"

So T=32-3L=20 (deg)

Solve the equation for L

-3L=20-32

-3L=-12

L=-12/-3=4

Subsituite L=4 in Volume equation

V=24-4(4) = 8 

Again, the strategy I used earlier above would save a lot of time but who knows what they would put in the exam.. It wouldn't work for far numbers apart. So Using arithmetic sequence is better.  

There might be a better answer. That's just my answer. 

Hope that helped :).

Tom has 6 red marbles, 4 blue marbles, and 3 green marbles.

How many distinct ways can he line them up in a row?

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{(6+4+3)!}{6!4!3!}} \\\\ &=& \dfrac{13!}{6!4!3!} \\\\ &=& \dfrac{6!\cdot 7\cdot 8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13}{6!4!3!} \\\\ &=& \dfrac{7\cdot 8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13}{4!3!} \\\\ &=& \dfrac{7\cdot 8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13}{1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 2 \cdot 3} \\\\ &=& 7\cdot 3\cdot 5\cdot 11\cdot 4\cdot 13 \\ &=& \mathbf{60060} \\ \hline \end{array}\)

\(\text{Applying Vieta's equations we have}\\ -7 = -(r_1+r_2+r_3)\\ \text{As we are told the roots are distinct positive integers the only possible solution is}\\ r_1=1,~r_2=2,~r_3=4,~\text{or some permutation thereof}\\ \text{Again using Vieta's equations (which return the same answer given all permutations)}\\ k = 1\cdot 2 + 1\cdot 4 + 2\cdot 4 = 14\\ m = 1\cdot 2\cdot 4 = 8\\ k+m=22\)

1)
Suppose the function \(f(x,y,z)=xyz\)
is defined for
\(x+y+z=7, x,y,z\geq 0\).
What is the range of \(f(x,y,z)\)?

\(\begin{array}{|rcll|} \hline f(x,y,z) &=& xyz \quad | \quad x+y+z=7 \text{ or } z = 7-x-y \\ &=& xy(7-x-y) \\\\ f(x,y) &=& xy(7-x-y) \\ f(x,y) &=& 7xy-x^2y-xy^2 \\ \hline f_x = \dfrac{\partial f(x,y)}{\partial x} &=& 7y-2xy-y^2 \\ f_y = \dfrac{\partial f(x,y)}{\partial y} &=& 7x-x^2-2xy \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & f_x=0 &=& 7y-2xy-y^2 \\ (2) & f_y=0 &=& 7x-x^2-2xy \\ \hline (1) & y(7-2x-y) &=& 0 \\ & \mathbf{y} &=& \mathbf{0} \\\\ & 7-2x-y &=& 0 \\ & \mathbf{y} &=& \mathbf{7-2x} \\ \hline (2) & x(7-x-2y) &=& 0 \\ & \mathbf{x} &=& \mathbf{0} \\\\ & 7-x-2y &=& 0 \\ & 7-x-2(y) &=& 0 \\ & 7-x-2(0) &=& 0 \\ & 7-x &=& 0 \\ & \mathbf{x} &=& \mathbf{7} \quad | \quad y=0 \\\\ & 7-x-2(y) &=& 0 \\ & 7-x-2(7-2x) &=& 0 \\ & 7-x-14+4x &=& 0 \\ & 3x &=& 7 \\ & \mathbf{x} &=& \mathbf{\dfrac{7}{3} } \quad | \quad y=7-2x \\\\ \hline & y &=& 7-2x \\ & y &=& 7-2(0) \\ & \mathbf{y} &=& \mathbf{7} \quad | \quad x=0 \\\\ & y &=& 7-2x \\ & y &=& 7-2(7) \\ & \mathbf{y} &=& \mathbf{-7} \quad | \quad x=7 \\\\ & y &=& 7-2x \\ & y &=& 7-2(\dfrac{7}{3}) \\ & \mathbf{y} &=& \mathbf{\dfrac{7}{3} }\quad | \quad x=\dfrac{7}{3} \\ \hline \end{array}\)

\(\begin{array}{|c|c|c|c|c|r|} \hline x&y&z=7-x-y & \text{solution} & f(x,y,z) & \\ \hline 0 &7 &0&&0&\text{minimum} \\ 7 &0 &0&&0&\text{minimum} \\ 7 &-7 &0& \text{no }(y\geq 0!)&& \\ \dfrac{7}{3} & \dfrac{7}{3} & \dfrac{7}{3} && \dfrac{343}{27} & \text{maximum} \\ \hline \end{array}\)

Range of \(f(x,y,z):\ 0\ldots \dfrac{343}{27}\)

Not sure if my answer is correct.

If sin(x)=2/3 

Therefore,

X=sin^-1(2/3)

=41.81 degrees.(Approx.)

If sec(y)=5/4

Therefore, y=sec^-1(5/4)=36.84 degrees.(Approx.)

So we know now that

X=41.81 degrees

Y=36.841

Using addition formula for the sin.

here:

Sin(A+B)=Sin(A)*Cos(B)+Sin(B)*cos(A)

Let X=A , B=Y

Sin(41.81+36.841)=Sin(41.81)*Cos(36.841)+Sin(36.841)*Cos(41.81)=56.175 (degrees) =0.98 (Radians)

Let me know if I did any mistake. 

\(\text{first place the red marbles, then the blue, the green go into the remaining slots}\\ N=\dbinom{13}{6}\dbinom{7}{4} = 60060\)

Thank you for the fast responce. THX

Principal  + Simple Interest  = 

10000 +  10000 ( interest rate) ( time in years)  =

10000  + 10000(.10)(5)  =

10000 + 5000 =

$`15000  after five years