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click a number and then hit divide or press the / key

At Jessie's fruit stall, 280 apples were sold in the morning and 3/5 of the remaining apples were sold in the afternoon. As a result, the number of apples left became 1/6 of the total apples at first. Given that the apples were sold at $2.50 in packs of six apples, how much did Jessie receive from the sale of the apples in the afternoon?

Let the original number of apples=A

Remaining apples after sale of 280 =A - 280

Remaining apples after the sale of 3/5=2/5[A - 280]

2/5[A - 280] = 1/6A, solve for A

A = 480 original number of apples.

3/5[480 - 280] =120 apples were sold in the afternoon.

120/ 6 = 20 packs of apples sold in the afternoon.

20 x $2.50 = $50 Total amount that Jessie recieved in the afternoon.

"At Jessie's fruit stall, 280 apples were sold in the morning and 3/5 of the remaining apples were sold in the afternoon. As a result, the number of apples left became 1/6 of the total apples at first. Given that the apples were sold at $2.50 in packs of six apples, how much did Jessie receive from the sale of the apples in the afternoon?"

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Thank you, 

r = sqrt [ (-6)^2 + (1)^2 ]    =  sqrt (37)

To find theta, we have

atan ( 1/ -6)   = -0.165148677415

However.....this point lies in the second quadrant so we must add pi radians to it.....and we have

-0.165148677415  + pi   ≈ 2.9764    rads

So   (-6, 1)  rectangular  =

(r, theta)   = ( sqrt (37),  2.9764)

Here in Portland Oregon we have finals on monday-friday. Then we get spring break. PLANET EARTH II is amazing, it took them 10 years to compile those videos. People need to watch how life really works, most are oblivious to it.

That's really specific....but sure why not 

also I got my phone taken away cause my BFF CALLED ME ON MY HOUSE PHONE AT MIDNIGHT AND IT WOKE EVERYBODY UP. When we get back to school ima k**l her

Read a book about the roman empire. Or even better go online and watch Planet Earth II.

3 + 1/2  =

7/2 =

3.5

650/1430 = 320/x   cross-multiply

650x   =  1430 * 320        divide both sides by 650

x =  ( 1430 * 320 ) / 650   =  704

u = < - 2, 8 >

v = < -7, - 9 >

cos (theta)   =   u (dot) v / [ ll u ll * ll v ll ]

u (dot) v   =   14 - 72  =   -58

ll u ll  =   sqrt [ (-2)^2 + (8)^2 ]  =  sqrt (68)

ll v ll   =   sqrt [ (-7)^2 + (-9)^2 ]   =  sqrt [ 49 + 81]   =  sqrt (130)

cos (theta)   =   -58 / [ sqrt (68) * sqrt (130) ]

arcos  [ -58 / [ sqrt (68) * sqrt (130) ] ]   = theta   ≈ 128.0888°

Here is a graph that shows this, Veteran :

here is another answer i got 

156.161259816828

u =  < -19. - 13 >

v = < 6 , -13 >

u + v    =   < -19 + 6, -13 + - 13 >    =  < -13, -26 >

ll u + v ll   =    sqrt [ (-13)^2 + (-26)^2 ]   ≈ 29.0689

Assuming this is the question:

(−6x² − 10) + (5x² + 6x − 5)

We can get rid of the parenthesees since only addition is happening between them.

−6x² − 10 + 5x² + 6x − 5

Now just combine like terms. We can rearrange it first to make it easier to see.

−6x² + 5x² + 6x − 10 − 5

−x² + 6x − 15

is anyone able to help me?  I need a formula to help me get these answers from the example

The easiest way is to look at a unit circle.

But not all angles have a rational sin or cosine.

The only ones that are on the unit circle (that I know of!) are 0º, 30º, 45º, 60º, 90º, and those same ones "flipped over" into the second, third, and fourth quadrants.

If you don't have a unit circle you can find the values by just drawing a triangle and using what you know about trig and the pythagorean theorem to find the values.

To use your example, this is how sin60º can be found:

First just draw out what you know sin 60º is. That's the black triangle.

And since there are 180º in every triangle, the remaining angle is 30º.

Next draw a triangle identical to it flipped over.

All the angles are 60º! This makes an equilateral triangle.

Since it is equilateral, the bottom line is also 1.

The black horizontal line is half that: 1/2.

(which is the cosine of 60º, but we are looking for sin.)

Now just go back to the good ol' pythagorean theorem

to find the vertical leg of that right triangle, aka sin60º.

(1/2)² + (sin60º)² = 1²

sin60º = √(1 - 1/4)

sin60º = √(3/4)

sin60º = √(3) / 2

That was one of my answers, the other one i got was like 84 degrees, i missed one class and it was about this stuff, so thanks. I will most likely ask a few more questions tonight if im stuck. I appriecate all the help around here.

tan (theta) =  (7 / -8)  

arctan (7/-8) = theta  = -41.1859°

The terminal point of this vector lies in the second quadramt....so the angle is

180 - 41.1859  ≈ 138.81407° ≈  138. 8141°

And this is the angle ( theta ) that the vector makes with the positive x axis

How many "square feet" are in the pool cover?

area = π * radius²

radius = diameter / 2

radius = 18 feet / 2

radius = 9 feet

area = π * (9 feet)²

area = 81π square feet

So there are 81π square feet in the pool cover.

...

Now let me ask this...

If there are 2 apples in a basket,

and the total cost of the basket is $6,

how much does each apple cost individually?

$6 / 2 apples = $3 per apple

...

If there are 81π square feet in the pool cover,

and the total cost of the pool cover is $48.43,

how much does each square foot cost?

$48.43 / (81π) square feet ≈ $0.19 per square foot *thx EP

v =  < 1, - 1 >

w = < 1, 2  >

v1   = [ v (dot) w  ] / l w l^2  *  < w >

v(dot) w   =   1* 1 + 2* -1   =  1 - 2    =   -1

l w l  =  sqrt (1^2 + 5^2)   =  sqrt (5)           l w l ^2   = 5

So......

v1   =   -1 /  (5)  *  <1,  2 >     =   < -1/5, -2/ 5  >

v2  =  v - v1  =  < 1 - -1/5, -1 - - 2/5 >  =  < 1 + 1/5, -1 + 2/5 > =   < 6/5,-3/5 >

Check

Sum of v1 and v2    =  < -1/5 + 6/5, -2/5 + - 3/5 >   =  < 1  -1 >    = v

If the tangent of an angle is 1/6, there are a few ways to find the other trig values.  First, cot(alpha) = 1/tan(alpha) so 1 over 1/6 is 6.  The formula sin²(alpha) + cos²(alpha)=1 can be divided through by cos² to give tan²(alpha) + 1= sec²(alpha).  The problem with plugging the 1/6 into that is you do not know which part of the circle or graph it is in so sec(alpha) upon solving would be sqrt[1/36 + 1] = sqrt[37/36] = sqrt(37)/6.  It might be negative depending on the original angle but let's just assume it is positive for now.  That brings us to sin and cos.   Since sec(alpha) = 1/cos(alpha), it = the reciprocal of sqrt(37)/6 which is 6*sqrt(37)/37.  Finally plug that back into sin²(alpha) + cos²(alpha)=1 to solve for cos(alpha).  

To summarize, use cot(alpha) = 1/tan(alpha) to get cot(alpha).  Use sec(alpha)= sqrt(tan²(alpha)+1) and then use that to get cos(alpha) since cos(alpha) = 1/sec(alpha).  Finally, use sin²(alpha) + cos²(alpha)=1 to get sin(alpha).  Since tan is positive in both quadrants 1 and 3, the sign on some of the trig functions might differ.

tangent = opposite / adjacent

a² + b² = c²

6² + 1² = (hypotenuse)²

√(37) = hypotenuse

sec = hypotenuse / adjacent

sec = √(37) / 6 ≈ 1.014

cos = adjacent / hypotenuse

cos = 6/√(37)

cos = [6√(37)] / 37 ≈ 0.986

cot = adjacent / opposite

cot = 6 / 1 = 6

sin = opposite / hypotenuse

sin = 1 / √(37)

sin = √(37) / 37 ≈ 0.164

csc = hypotenuse / opposite

csc = √(37) / 1 = √(37) ≈ 6.083

Go online to this page and learn the relationship between the 6 Trig. functions:

http://www.efunda.com/math/trig_functions/trig_relation.cfm

Find the least value of b such that b²+ 2b - 15 ≤ 0 .

First let's find what value of b makes

b²+ 2b - 15 = 0

Factor it.

(b - 3)(b + 5) = 0

Set each factor equal to zero.

b - 3 = 0          and          b + 5 = 0

b = 3               and          b = -5

Now to find out what all values of b makes it less than 0,

let's test a point in the middle of -5 and 3, say 0, and see if that makes the inequality true.

0² + 2(0) - 15 ≤ 0

- 15 ≤ 0  true

Just to be safe let's test a point less than -5, say -6, and a point greater than 3, say 4.

36 - 12 - 15 ≤ 0

9 ≤ 0  false

16 + 8 - 15 ≤ 0

9 ≤ 0  false

So all this means that -5 ≤ b ≤ 3

\(\frac{\pm 3}{150}*100 \\=\frac{\pm 300}{150} \\= \pm 2 \\~\\ 2 \hspace{1cm} \text{or} \hspace{1cm} -2\)

Unless if you meant:

\(\frac{\pm 3}{150*100} \\ = \frac{\pm 3}{15,000} \\ =\frac{\pm 1}{5,000} \\~\\ \frac{1}{5,000}\hspace{1cm}\text{or}\hspace{1cm} -\frac{1}{5,000}\)