All Answers 358180 Answers

Here's # 2.......we need to manipulate the Law of Cosines twice.....

We can first  find  angle RST...so we have

[ RS^2  +ST^2 - RT^2] / [ 2(RS)(ST)] = cos RST

So

[13^2 + 14^2  - 15^2] / [ 2(13) (14) ]  =  cos RST

So  

arcos  ( [13^2 + 14^2  - 15^2] / [ 2(13) (14) ] )  =  RST  ≈  67.38°

Since M is the mid-point of ST, then SM  = 7

Now manipulating the Law of Cosines again to find RM, we have that

RM  =  sqrt ( SM^2  + RS^2  - 2(SM)(RS) cos  67.38° )  

RM =  sqrt ( 7^2  + 13^2 - 2 (7)(13) cos  67.38° ) ≈ 12.166

Here's an approximate pic :

1 - 

Verify the following identity:
(cos(θ) + 1) (1 - cos(θ)) = sin(θ)^2

(1 - cos(θ)) (cos(θ) + 1) = 1 - cos(θ)^2:
1 - cos(θ)^2 = ^?sin(θ)^2

sin(θ)^2 = 1 - cos(θ)^2:
1 - cos(θ)^2 = ^?1 - cos(θ)^2

The left hand side and right hand side are identical:
Answer: | (identity has been verified)

2-

Verify the following identity:
(sin(θ))/(cos(θ)) + (cos(θ))/(sin(θ)) = csc(θ) sec(θ)

Put (cos(θ))/(sin(θ)) + (sin(θ))/(cos(θ)) over the common denominator sin(θ) cos(θ): (cos(θ))/(sin(θ)) + (sin(θ))/(cos(θ)) = (cos(θ)^2 + sin(θ)^2)/(cos(θ) sin(θ)):
(cos(θ)^2 + sin(θ)^2)/(cos(θ) sin(θ)) = ^?csc(θ) sec(θ)

Multiply both sides by sin(θ) cos(θ):
cos(θ)^2 + sin(θ)^2 = ^?cos(θ) csc(θ) sec(θ) sin(θ)

Write cosecant as 1/sine and secant as 1/cosine:
cos(θ)^2 + sin(θ)^2 = ^?1/(cos(θ)) 1/(sin(θ)) cos(θ) sin(θ)

cos(θ) (1/(sin(θ))) (1/(cos(θ))) sin(θ) = 1:
cos(θ)^2 + sin(θ)^2 = ^?1

Substitute cos(θ)^2 + sin(θ)^2 = 1:
1 = ^?1

The left hand side and right hand side are identical:
Answer: | (identity has been verified)

Is this what you are looking for?

e^ix = cos(x) + i*sin(x)      (where i = square root of -1)

.

Easiest way to see this is to look at some numerical examples.  In the matrices below you can see that the LHS = RHS for all values of m for x₁, but LHS = RHS only for odd values of m for x₂.

(The first column of a matrix below is m, the second is the LHS of your original equation, the third is the RHS)

Use this formula to calculate the required period:

FV=P{[1 + R]^N - 1/ R}

25,000 = 245 x {[ 1 + 0.057/12]^N - 1 / (0.057/12]}

25,000 = 245 x {[ 1 + 0.00475]^N - 1 / (0.00475)}

25,000 = 245 x {[1.00475]^N - 1 / (.00475)} divide both sides by 245

102.040816 ={[1.00475]^N - 1 / (0.00475)}

Solve for N:
102.041 = 210.526 (1.00475^N - 1)

102.041 = 210.526 (1.00475^N - 1) is equivalent to 210.526 (1.00475^N - 1) = 102.041:
210.526 (1.00475^N - 1) = 102.041

Divide both sides by 210.526:
1.00475^N - 1 = 0.484694

Add 1 to both sides:
1.00475^N = 1.48469

Take the logarithm base 1.00475 of both sides:
Answer: | N = 83.3993 =83.40 Months, or about 6 years and 11.4 months.

y² = 4

y=2

because you can take the square root from y² which gives you y and then take the square root of 4 which is equivalent to 2.

To answer the Guest's question, you pick three white balls and one blue Superball.

To answer Melody's question, you pick three numbers from 1 to 20. Once you pick a number from there, it can not be chosen again. You also pick a number from 21 to 30. You pick these numbers at random. ( I never played lottery before, so I'm doing my best.) Again, three numbers from 1 to 20 and one number from 21 to 30 is chosen. These numbers are the winning numbers. If it happens that you match at least two of the numbers from 1 to 20 with the winning numbers, or you match the Superball number, you win a prize. ( The order doesn't matter)

\((f\circ g)(-4)=f(\,g(-4)\,)\)

f(x)  =  4x + 7                             Replace every instance of   x   with   g(-4)   .

f( g(-4) )  =  4( g(-4) ) + 7

And.....

g(x)  =  3x - 5                             Replace every instance of   x   with   -4   .

g(-4)  =  3(-4) - 5                        Simplify.

g(-4)  =  -17

f( g(-4) )  =  4( g(-4) ) + 7             On the right side, replace   g(-4)   with   -17

f( g(-4) )  =  4( -17 ) + 7               Simplify.

f( g(-4) )  =  -61

Y²  =  4                    Take the  ± square root  of both sides.

Y    =  ±√4

Y    =  ± 2

Thanks for your answer.

"However, x_2 is only valid when m is an odd integer."

This may be a silly question, but could you explain to me why that is the case? Thanks.

Hi Guest,

Thanks for your answer,

Why don't you start learning LaTex?

It is not that hard, there is always people here that will help you get started.

You can start just by using it for fractions and very basic things and progress from there.

To write   \(5\frac{2}{3}\)

You just open the LaTex tab in the ribbon. Delete the formula that is already there and type in

5\frac{2}{3}          and press  ok :)

You have still not explained how trhe game is played to us!!

Prove that sin(30°) = sin (390°)

I suppose that depends on what you are allowed to use for your proof.

How about

sin390 = sin(360+30) = sin360cos30+sin30cos360 = 0+sin30 =sin30

As follows:

.

I do not know how to solve the second problem; however, I can help solve the first problem.  To solve the first problem, use Heron's Formula.

\(A=\sqrt{\frac{a+b+c}{2}\times(\frac{a+b+c}{2}-a)\times(\frac{a+b+c}{2}-b)\times(\frac{a+b+c}{2}-c)}\)

A = Area

a = side a or \(\frac{4}{3}\)

b = side b or \(2\)

c = side c or \(\frac{8}{3}\)

\(A=\sqrt{\frac{\frac{4}{3}+2+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{\frac{\frac{10}{3}+\frac{8}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{\frac{\frac{18}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{\frac{\frac{18}{3}}{2}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{\frac{18}{3}}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{6}{2}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{18}{6}-\frac{4}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{18}{6}-\frac{8}{6})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{10}{6})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times(\frac{5}{3})\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{3\times\frac{5}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{\frac{3}{1}\times\frac{5}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{\frac{15}{3}\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{18}{3}}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{6}{2}-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(3-2)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(1)\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times1\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{4}{3}+2+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{4}{3}+\frac{6}{3}+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{10}{3}+\frac{8}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{\frac{18}{3}}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{6}{2}-\frac{8}{3})}\)

\(A=\sqrt{5\times(3-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{9}{3}-\frac{8}{3})}\)

\(A=\sqrt{5\times(\frac{1}{3})}\)

\(A=\sqrt{5\times\frac{1}{3}}\)

\(A=\sqrt{\frac{5}{1}\times\frac{1}{3}}\)

\(A=\sqrt{\frac{5}{3}}\)

Terence has asked me for more explanation so I shall explain some more:

(a) superior construction Pte Ltd is a successful company dealing with many major projects in singapore.

recently, it has summited its biddings for two major govt projects. projects A worth $120million and the company believes it has 40% chance of securing the projects. project B worth $1.8 billion and there is 30% chance the company can win the project.both project are independent of each other. what is the probability that the company:

The value of the projects is irrelevant to this question.

It is best to draw a venn diagram to answer questions like these. In this case that will be two overlaping circles representing success in secruing A and B projects and a surounding rectangle whit the probability of not getting either project. This will help you work out what you are bing asked for. I can draw it if you need me to but its a hassle and you can probably do it for yourself on a peice of paper.

We have

P( securing A) = 0.4    given

P( securing B) = 0.3     given

P(securing both)=0.3*0.4=0.12    If you wnat the probablility of 2 independent events happening then you muliply the probabilities together. 

Underneath in the box is supposed to be 'neither A NOR B'   This is what is outside the circles. 

P(securing A but not B) = 0.4-0.12 = 0.28

P(securing B but not A)= 0.3-0.12 = 0.18

P(securing either) = 0.4+0.3-0.12 = 0.58       This includes the prob of securing both

You can see from the diagram that the 0.12 is included in both circles.  

If you do not subtract it here then you will have added it in TWICE. You don't want that :)

P(securing neither) = 1-P(securing either) = 1- 0.58 = 0.42

I've subtracted the prob of getting any contract from 1 because the possibility of anything at all happening is 1.

(i) will secure Project A or B but not both = 0.28+0.18 = 0.46

(ii) will not secure project A or will not secure project B =  1 - P(securing both) = 1-0.12 = 0.88

Does that help?

7x + 3 * (8 * 9)  =  1224                  We don't need to keep the parenthesees here.

7x + 3 * 8 * 9    =  1224                  Replace   3 * 8 * 9   with   216

7x + 216  =  1224                           Subtract   216   from both sides of this equation.

7x + 216 - 216  =  1224 - 216

7x  =  1008                                     Divide both sides of this equation by 7.

7x / 7  =  1008 / 7

x  =  144

Come on.. try to understand what I have written.

If you understand the significance of the   'co'  in the front then all this becomes dead easy.

tangent and cotangent is

tan and cot

so

tan(x) = cot(90-x)

If tan(x)=cot(x)

then

x must equal 90-x

etc!

Sin(θ)=cos (θ)

Tan (θ)=cot (θ)

sine and cosine

tangent and cotenagent.

see that co?  

That stands for complimentary .. and

Complimentary angles add to 90 degrees

so 

\(Sin(\theta)=cosine (90-\theta)\\ so\;\;if\\ Sin(\theta)=cosine (\theta)\;\;then\\ \theta = 90-\theta\\ 2\theta=90\\ \theta=45^\circ\\ \text{of course there are an infinite number of answers but they are all of the from}\\ \theta =45+2\pi n\qquad n\in Z \)

You can do the other one yourself :)

A amoeba travels 4.2*10^(-4) m in 30s. What is the speed in ms^(-1)? Give your answer in standard form

Equation:V(speed)= D(distance)/T(time)

V =4.2*10^(-4) / 30 =0.000014 ms^-1 - Amoeba's speed in meters per second.

Solve the linear equation y(x) + ( dy(x))/( dx) = x:
Let μ(x) = e^( integral1 dx) = e^x.
Multiply both sides by μ(x):
e^x ( dy(x))/( dx) + e^x y(x) = e^x x
Substitute e^x = ( d)/( dx)(e^x):
e^x ( dy(x))/( dx) + ( d)/( dx)(e^x) y(x) = e^x x
Apply the reverse product rule g ( df)/( dx) + f ( dg)/( dx) = ( d)/( dx)(f g) to the left-hand side:
( d)/( dx)(e^x y(x)) = e^x x
Integrate both sides with respect to x:
 integral( d)/( dx)(e^x y(x)) dx = integral e^x x dx
Evaluate the integrals:
e^x y(x) = e^x (x - 1) + c_1, where c_1 is an arbitrary constant.
Divide both sides by μ(x) = e^x:
Answer: | y(x) = x + c_1 e^(-x) - 1

Sin(390) = sin(360 + 30) = sin(30) = 1/2

Sin(360) =0

Sin(390) = sin(0 + 30) =sin(30) = 1/2

Go online and use this calculator to get the area of your triangle:

http://www.mathopenref.com/heronsformula.html

Tricky Mr. BB,

Presenting facts that doesn’t answer the question.

I can do that too: 2+2=4

-----

nCr(20, 2) = 190 ways of selecting 2 from 20

nCr(10, 1)    =   10 ways of selecting 1 from 10

To find the over all probability of winning,

Calculate the Harmonic Mean (19) and divide by 2

Equivalent to

1/ (1//190 + 1/10) =9.5

Over all odds are 1 in 9.5.