All Answers 352710 Answers

ok thanks.

Are you doing a unit that might give us a hint of what you are expected to do here?

Have you done any calculus yet?

nope, it's x^2 and 16/x added together, no brackets.

Thank you   

I actually get annoyed pretty easily    

He was just some lazy, and rather stupid, teenager using the site to cheat. 

He was an instant reincarnation of Bradv23. That is how I know he is stupid as well as dishonest and lazy.

To figure this out, we need to figure out the area of the pink parellelogram.

Evaluate 6*5 to get 30 square (units).

*The area of a trapezoid is ((a+b)/2)*h.

* =  When I say a and b, those are the 3 and 7 side lengths.

Now, we can say the area is 5*h, or 5h. 

Because the area of the parallelogram is 30 and the area is the same for both shapes, 5h must equal 30.

Therefore, h or height is 6.

~~Hypotenuisance

Let C  be the cost per student before the price reduction....then we have that

20C  =  (20 + 10) (C - 1.50)     simplify

20C  = 30 ( C - 1.50)

20C  =  30C -  45      add 45 to both sides, subtract 20C from both sides

45 = 10C        divide both sides by 10

4.5  = C  = $4.50

So...the cost for the bus is

20 (4.5)  =

$90

OK....you got one angle

Since the tangent is also positive in the 3rd quadrant, add this reference angle to 180°  to get the other angle =

259.6°

And that's all of them  on [ 0, 360 ]    !!!!

LOL! idk what the question was but it must have been really bad to p**s off melody. BTW-happy 100k points!

Is the x^2 + 16 supposed to be in brakets?

We have the following inequalities

8 + 15 >  x        where x is the unknown side

23 > x

x < 23         and

8 + x >  15

x > 7

However....since the triangle is acute....we know that  an 8 - 15 - 17  triangle is a right triangle

So....the remaining side, x, must be <  17  [  any integer > 17  but < 23  will produce an obtuse triangle ]

So....the possible side lengths are integers  between 7  and  17

However....we must even restrict this interval  to  integers  greater than 12 and < 17 

The reason for this is that integer side lengths for x  from 8 to 12 inclusive also produce obtuse triangles 

So....4 possible acute triangles

Solve for x :
3^(x^2 + 4 x + 4) = 9^(x + 2)

Take the natural logarithm of both sides and use the identity log(a^b) = b log(a):
log(3) (x^2 + 4 x + 4) = 2 log(3) (x + 2)

Divide both sides by log(3):
x^2 + 4 x + 4 = 2 (x + 2)

Expand out terms of the right hand side:
x^2 + 4 x + 4 = 2 x + 4

Subtract 2 x + 4 from both sides:
x^2 + 2 x = 0

Factor x from the left hand side:
x (x + 2) = 0

Split into two equations:
x = 0 or x + 2 = 0
Subtract 2 from both sides:
x = 0     or     x = -2

Solve for x over the real numbers:
abs(2 x^2 + 3 x - 1)^(3 x^2 - 13 x + 4) = 1

Take the natural logarithm of both sides:
log(abs(2 x^2 + 3 x - 1)) (3 x^2 - 13 x + 4) = 0

The left hand side factors into a product with three terms:
log(abs(2 x^2 + 3 x - 1)) (x - 4) (3 x - 1) = 0

Split (x - 4) (3 x - 1) log(abs(2 x^2 + 3 x - 1)) into separate parts with additional assumptions.
Assume abs(2 x^2 + 3 x - 1)!=0 from log(abs(2 x^2 + 3 x - 1)):
x - 4 = 0 for abs(2 x^2 + 3 x - 1)!=0
 or 3 x - 1 = 0 for abs(2 x^2 + 3 x - 1)!=0
 or log(abs(2 x^2 + 3 x - 1)) = 0

Add 4 to both sides:
x = 4 or 3 x - 1 = 0 for abs(2 x^2 + 3 x - 1)!=0
 or log(abs(2 x^2 + 3 x - 1)) = 0

Add 1 to both sides:
x = 4 or 3 x = 1 for abs(2 x^2 + 3 x - 1)!=0
 or log(abs(2 x^2 + 3 x - 1)) = 0

Divide both sides by 3:
x = 4 or x = 1/3 or log(abs(2 x^2 + 3 x - 1)) = 0
Cancel logarithms by taking exp of both sides:
x = 4 or x = 1/3 or abs(2 x^2 + 3 x - 1) = 1

Split the equation into two possible cases:
x = 4 or x = 1/3 or 2 x^2 + 3 x - 1 = 1 or 2 x^2 + 3 x - 1 = -1

Divide both sides by 2:
x = 4 or x = 1/3 or x^2 + (3 x)/2 - 1/2 = 1/2 or 2 x^2 + 3 x - 1 = -1

Add 1/2 to both sides:
x = 4 or x = 1/3 or x^2 + (3 x)/2 = 1 or 2 x^2 + 3 x - 1 = -1

Add 9/16 to both sides:
x = 4 or x = 1/3 or x^2 + (3 x)/2 + 9/16 = 25/16 or 2 x^2 + 3 x - 1 = -1

Write the left hand side as a square:
x = 4 or x = 1/3 or (x + 3/4)^2 = 25/16 or 2 x^2 + 3 x - 1 = -1

Take the square root of both sides:
x = 4 or x = 1/3 or x + 3/4 = 5/4 or x + 3/4 = -5/4 or 2 x^2 + 3 x - 1 = -1

Subtract 3/4 from both sides:
x = 4 or x = 1/3 or x = 1/2 or x + 3/4 = -5/4 or 2 x^2 + 3 x - 1 = -1

Subtract 3/4 from both sides:
x = 4 or x = 1/3 or x = 1/2 or x = -2 or 2 x^2 + 3 x - 1 = -1

Divide both sides by 2:
x = 4 or x = 1/3 or x = 1/2 or x = -2 or x^2 + (3 x)/2 - 1/2 = -1/2

Add 1/2 to both sides:
x = 4 or x = 1/3 or x = 1/2 or x = -2 or x^2 + (3 x)/2 = 0

Factor x and constant terms from the left hand side:
x = 4 or x = 1/3 or x = 1/2 or x = -2 or 1/2 x (2 x + 3) = 0

Multiply both sides by 2:
x = 4 or x = 1/3 or x = 1/2 or x = -2 or x (2 x + 3) = 0

Split into two equations:
x = 4 or x = 1/3 or x = 1/2 or x = -2 or x = 0 or 2 x + 3 = 0
Subtract 3 from both sides:
x = 4 or x = 1/3 or x = 1/2 or x = -2 or x = 0 or 2 x = -3

Divide both sides by 2:
x = 4     or x = 1/3     or x = 1/2     or x = -2     or x = 0     or x = -3/2

Hi Hearts123,

Chris has answered it using calculus.   Have you done any calculus yet?

That is the normal method I would use too.

[The differential of x^2 = 2x]

BUT

I do not think that is what you want.

The trouble is that neither of us have worked out the relevance of that inequality (which is always true by the way).

I will think about it some more :)

A and B are constants such that the graphs of the lines 3x - 4y = 7 and 8x + Ay = B are perpendicular and intersect at (5,2). What is A+B?

Slope   of first line  =    3/4

Slope of second line = -4/3

Rearrange the second equation

Ay  = -8x+ B

y  =  (-8/A)x + B/A

Which implies that  -8/A =  -4/3   so    24 = 4A  ⇒  A = 6

And since (5,2) is on the both lines...then

(6) 2  = -8(5) + B

12 = -40 + B

52  = B

A + B  =   6 + 52  =  58

Here's a graph : https://www.desmos.com/calculator/4q0eh6wjme

thank you so much, I really appreciate it :)

I'm just wondering, was the mean inequality used for what you just did? 

And why did the x^2 turn into 2x?

Kawaiiai,

I just looked at your profile page.

Why don't you EVER say thank you to ANYONE! 

Saying it in advance is not be a replacment for a proper thank you when help is given to you.

Giving points is good but they should be additional to a thank you.

If you do not respond to my criticisms there may be consequences for you.

y  =  x^2 + 16x^(-1)

Take the derivative....set to 0  and solve

y '  =   2x  - 16/x^2  =  0

2x^3 - 16  = 0

x^3 - 8  = 0

Taking the positive cube root produces  x  = 2

And the minimum value  is     ( 2)^2  + 16/2   =   4  + 8 =   12

(a)

Prob 0 orange   = (2/8)^2  = (1/4)^3  = .0156

Prob 1 orange  =  C(3,1) * (1/4)^2 * (6/8)  = .1406

Prob 2 orange = C(3,2) *(1/4) * (3/4)^2  = .4219

Prob 3 orange  = (3/4)^3  = .4219

3rd histogram

(b)

Expected number of orange = 0 + .1406 + 2(.4219) + 3(.4219) = 2.25

Prob 0 green balls  = (3/7)^3  = .0787

Prob 1 green =   C(3,1) * (4/7) * (3/7)^2  = .3149

Prob 2 green  = C(3,2) * ( 4/7)^2 * (3/7) =  .4198

Prob 3 green =  (4/7)^3  = .1866

5 min 36 sec  =   5 + 36/60  =  5 + 3/5  =  5.6 minutes

So

5 * [2 pi (4.35)]  (5.6)  ≈  765.3  cm

kawaiiai,

Are you too lazy to scan through a short internet page to find the answer you are looking for.  Even when someone else has given you a direct link?

How do I know your displeasure?    That is easy.     I can see how you allocate your points. 

You should be very grateful that someone is prepared to provide you with the internet links that will help you.  I guess you are not looking for long term help.

Like many people here.  Your attitude is     "JUST GIVE ME THE FRICK'N ANSWER!"

Well some others here are always happy to oblige.  Arn't you lucky.

PEMDAS - Paranthesis Exponents Multiplication Division Addition Subtraction. Do it that order.

1659...

One thousand six hundred fifty-nine.