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D (1, 3), E (3, 2), F (1, -1), G (-3, -2), and H (-2, 2), which statement is true about figure D'E'F'G'H'? 

D (1, 3),  1st quad goes to 4th quad D'(3,-1)      swap numbers, x is pos and y is neg

E (3, 2),      As above  E'(2-3)

F (1, -1),    the quad goes to 3rd quad   swap the abs numbers, x is neg y is neg    F'(-1,-1)

G (-3, -2)    third quad goes to 2nd quad  swap the numbers , x is neg y is pos  G'(-2,3)

H (-2, 2), second quad goes to 1st quad,  swap the numbers,  x is pos and y is pos    H'(2,2)

Why don't you join up properly - there are lots of benefits and no disadvanges - it is really easy.  :)

$$\\\frac{\$AU\;1}{\$NZ\;0.931}\times \frac{\$NZ\;788}{1}=\frac{\$AU788}{0.9}\\\\
$ the NZ dollars cancel out$$$

15" and you can google the conversion.      It is times about 2.5 I think.

I would have just multiplied both sides by 8 from the very beginning and gotten rid of all the fractions. :)

x + 3/2 = 3/4 (x - 1/2)   →

(1/4)x = -3/8 - 3/2

(1/4)x  = -(6 + 24)/16

4x = -30

x = -7.5

I do not thinkt that this is true  

Essentially this is the question

prove

$$\dfrac{v^2sin^2\alpha}{2g}=\dfrac{-g}{2}\left(v^2\times\dfrac{sin\alpha}{g}\right)^2+(vsin\alpha)\left(\dfrac{vsin\alpha}{g}\right)\\\\\\
\begin{array}{rll}
RHS&=&\frac{-g}{2}\left(v^4\times\frac{sin^2\alpha}{g^2}\right)+\left(\frac{v^2sin^2\alpha}{g}\right)\\\\
&=&\frac{-1}{2}\left(\frac{v^4sin^2\alpha}{g}\right)+\left(\frac{v^2sin^2\alpha}{g}\right)\\\\
&=&\left(\frac{-v^4sin^2\alpha}{2g}\right)+\left(\frac{2v^2sin^2\alpha}{2g}\right)\\\\
&=&\frac{v^2sin^2\alpha}{2g}\;(-v^2+2)\\\\
&=&\frac{v^2sin^2\alpha\;(2-v^2)}{2g}\;\\\\
\end{array}\\\\
$ This does not equal the LHS $$$

sry my mistake

x+3/2=3/4x-3/8

-3/4x+x==3/2-3/8

-1/4x=-15/8

x=8/60

x=0.1333333333333333

x+3.2=3.4x-4.08

-3.4x+x=-3.2-4.08

-2.4x=-7.28

x=-7.28/-2.4

x=3.03333333333333

Let x be your normal hourly rate....So, you work 40 hrs at the rate of x, and 5 more hrs. at 1.5x....so we have

40x + (5)*(1.5x) = 475  simplify

40x + 7.5x = 475

47.5x = 475      divide both sides by 47.5

x = $10/hr

It is just a function.  It is not a question.  

How do I calculate this? tan x=-2. sin x=? What is the formula?

To start with x muxt be  in the 2nd or 4th quad because tanx is neg

this means that sinx can be neg or pos

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

Now forget about the negative for the moment

Draw a right angled triangle.

Put x in one of the acute angles

tanx=2/1   so put 2 on the opposite side and 1 on the adjacent side.

Use Pythagoras' theorum to get the hypotenuse.   $${\sqrt{{\mathtt{5}}}}$$

So reading off the triangle  $$sinx=\pm\frac{2}{\sqrt5}$$

sin(2a) = 2sin(a)cos(a)

sin(a)sin(a) = sin²(a)

6y-3(y-2)     there's nothing here to solve...just to simplify....distribute the -3 over the terms in the parentheses

6y - 3y + 6 =

3y + 6

You have an odd number of brackets  but

$$\\-5^2\times(1-2(-2)\div -1)\\\\
=-25\times(1+4\div -1)\\\\
=-25\times(1+-4)\\\\
=-25\times(-3)\\\\
=75\\\\$$

If tan x=-2, then either  y = 2 and x = -1, or y= -2 and x = 1.

And regardless, r = √5   ...... And sinx = y/r

So, either sin x = 2/√5  and x is a 2nd quadrant angle

Or, sin x = -2/√5     and x is a 4th quadrant angle 

Hi Ninja and Golden Leaf.  It is great to see both of you :)

I love your hello pic Ninja.    

(-3/5)x>18   multiply both sides by 5 to clear the fraction

-3x > 90       divide both sides by -3 and remember that we have to reverse the inequality signs when we divide by a negative...

x < -30

I don't think your answer is correct....rotating (-2,2) 90 degrees clockwise puts us in the 1st quadrant. And since H is already in the 2nd quadrant at (-2, 2), it can't remain in the 2nd quadrant at (-2,3).  The likely candidate for the correct answer is A. Follow the logic...If E is in the 1st quadrant, its x and y coordinates are positive. Then, rotating it 90 degrees clockwise would put it in the 4th quadrant....Hence, the x coordinate would be positive and the y coordinate would be negative....  the rotational matrix for this is given by

|   cos90    sin90  |   

| -sin90    cos 90  |

So we have

|   cos90    sin90  |    x  l 3 l

| -sin90    cos 90  |       l 2 l        .....  and this gives.......

(3 cos 90 + 2 sin 90) = ( 0 + 2) = 2 ......for the resulting x  coordinate

And ( -3sin 90 + 2cos 90)= (-3 + 0 )  = -3 .......for the resulting y coordinate

Yup...just as I suspected !!!.....the resulting point is (2. -3) and "A" is the correct choice.....

Thank you!

Let s+1 = u

Then s = u -1

And ds = 1 du

So we have

4pi∫(u - 1)/u³ du =

4pi [∫1/u^-2 du - ∫ u^-3 du ] + C =

4pi [ (-1)u^-1 - (-1/2)u^-2 ]  + C =

4pi [ 1/(2u²) - 1/u ] + C =

2pi [ 1- 2u]/ u² + C =   [back substitute ....( u = s+ 1) ]

2pi [ 1 -2(s + 1) ]/ (s + 1)² + C

2pi [ -(2s + 1) ] / (s + 1)² + C =

-2pi (2s + 1) / (s + 1)² + C

The probability of at least one red one is...... 1 - P(no red selected). In essence, we're just counting the number of possible sets of green marbles only divided by the number of total possible sets and subtracting this result from 1.

So we have......

1 - C(6,3) / C(10, 3)  = 5/6 or about 83.333%

first expand the brackets on the left hand side to:

9p-36=5p+4

then put all the p's on one side and all the numbers on the other:

9p-5p=4+36

simplify:

4p=40

Divide the right hand side by 4:

p=40/4

so:

p=10

Hah, me too. For me life is going well, and I hope it is for you, too! ^^

Thanks for that answer, Alvarez......nicely presented......!!!