All Answers 358097 Answers

a + ab^2  = 40b

a - ab^2  = -32b       add these

2a  = 8b        divide by 2

a  = 4b    (1)

Sub (1) into either equation

(4b) + (4b)b^2  = 40b

4b^3 + 4b - 40b   = 0   

4b^3 - 36b = 0

4b ( b^2 - 9)  = 0

4b ( b + 3) (b - 3)  = 0      set each factor to 0  and solve for b and we have that

b =  0        b  = - 3        and b  =  3

And a  = 4(0)  = 0         4(-3)  = - 12        4(3)  = 12

So....the solutions are   (a, b)  = (0, 0)    (-3, -12)    and  ( 3, 12)

Distance from A to C  =  sqrt [ (1- 4)^2 + (1 - 0)^2 ]  =  sqrt [ (-3)^2 + (1)^2 ]  = sqrt (10)

Distance from A to B  =  sqrt [ (1 - 2)^2 + (1 - 3)^2 ]  =  sqrt [ (-1)^2 + (-2)^2 ]  =  sqrt (5)

Distance fromB to C =  sqrt [ (4 - 2)^2 + (3 - 0)^2 ]  =  sqrt [ 2^2 + 3^2 ]  =  sqrt [13]

So........using the Law of Cosines we have that

[sqrt (13)]^2  - [sqrt(10)]^2 - [sqrt (5)]^2   

________________________________  = cos theta

     -  2 sqrt (10)*sqrt(5)

[ 13 - 10 - 5 ]

__________   =   cos theta

-2 sqrt (50)

[-2]

________    = cos theta

-2 sqrt (50)

1

_______    =  cos theta

sqrt (50)

1

_______   =  cos theta

5 sqrt (2)

So  x  =  1/5

9(x - 1)^2 +  y^2  = 36      divide both sides by 36

(x - 1)^2          y^2

______   +      ____     =  1

    4                   36

The center is  ( 1, 0)

The major axis is along y   and the minor along x

One of the enpoints on the major axis is  (1, 0 + 6)   =  ( 1, 6)

One of the points on the minoraxis is (1 + 2, 0)  =  (3, 0)

So the distance between these points is   sqrt [ ( 3 - 1)^2 + ( 6 - 0)^2 ]  =  sqrt [ 2^2 + 6^2 ]  =

sqrt [40]  =

2sqrt(10)  units

Here's a graph :  https://www.desmos.com/calculator/2zr89eh7yu

Yes I understand that but I like that you are giving answers with your code that so far have shown to be correct.

It is like being given the answer from the back of the text book. I like that you are doing it.

Coding is something that I would like to study but there are so many things I want to study that I do not know if I wil ever get to it.

I have forgotten most of what I learned in the Jurassic period too  

GPA (Grade point average) is worked out differently in different countries, and you have not stated your country.

High and low are also interpreted differently by different people/situations

For example, are you aiming at being at being a doctor or a nurse. 

Cool....I was just pointing out that people who have taken AP classes could have a 4.4 GPA   (or perhaps higher).

I understand. I'm not taking any AP classes. This is my senior year and I only have three courses, Sociology II, Practical Math, and British World Literature is all. Thanks for your advice. :)

Usually a GPA only goes as high as 4......in recent years , with AP classes giving an extra point for each letter grade, GPAs can be higher than 4.  It is hard to say what your GPA is worth without knowing the entire range of to whom you are comparing yours.

no prob

Okay, thanks so much

It is a good gpa because if you had less than 3 it would be bad and if you had 4 or higher it would be really good.

First, calculate the equation of the dotted line using two convenient points (  1,-2    and   3,1 )     then the yellow portion is

y >  equation of the  line

Did anyone ever get the answer to this?

\(f(x) \text{ is even, so }f(-x)=f(x)\)

\(g(2)=f(2+3)-5 = f(5)-5=-2\\ f(-5)=f(5)\\ f(-5)-5=-2\\ f(-8+3)-5 = -2\\ g(-8)=-2\\ (a,b)=(-8,-2)\)

A quadrilateral with two pair of adjacent equal length sides.

a quadriateral with two sets of adjacent sides having equal length.

Think of the usual example of a kite.

Use right triangle trig properties    SOHCAHTOA

One leg is 90  (this is the ADJACENT leg to the 48 degrees)

The 'straight line' will be the hypotenuse of the right triangle and the distance you are looking for.....

cos 48 = adj  /hyp

cos 48 = 90/hyp         Solve for hyp.

(Thanx, Chris for pointing out my error......I was reading 90 ft as the height of the cliff.....it is the ditance to the base of the cliff!  D'Oh!)  

There are 54 coins     Let x be the number of quarters.....then 54-x is the number of nickles

.25 x = value of quarters

.05 (54-x) = value of nickles.....       added together, these must equal 9.10

.25x   +  .05(54-x) = 9.10       solve for 'x',,,the number of quarters, the rest will be nickles (54-x)

Go !  

All the choices are off by a factor of 5 or more just based on the volume of the warehouse and the volume of the cylinders

Opposite divided by hypotenuse is the sine of the angle, so you need to find the arcsine (asin, sin^-1) of 7/12.

I'm unsure if I'm interpreting this correctly but here goes lmao.

\(\frac{71}{5} * (6^7 - (-986533) * 3268358218) * 6*555535352 + (55555*636218*368*96492) * 5\)

\(\frac{71}{5} * (279936 + 3224343237878194) * 3333212112 + 6275354076445027200\)

\(\frac{71}{5} * 3224343238158130 * 3333212112 + 6275354076445027200\)

\(152613363072370508719241952 + 6275354076445027200\)

\(152613369347724585164269152\)

The floor of 137/2 is 68.

So m=68.

You are very welcome!

:P

Hey, that's a good way to do it!

Another way of writing \(\lfloor x\rfloor = 4\) defines x values in the interval of \(4 \leq x < 5\)

If we multiply this interval by 4, we have that \(16\leq 4x < 20\).

In order to get the sum, we have to acknowledge that our 4x is still being floored, given \(\lfloor 4x \rfloor\). Therefore, only integer values within the interval can be a result of this. For example, \(x = 4.625\) satisfies that \(\lfloor x \rfloor = 4\). Then, \(\lfloor 4x \rfloor = \lfloor 4*4.625\rfloor = \lfloor 18.5 \rfloor = 18\)

The integers within \([16, 20)\) are 16, 17, 18, and 19. \(16+17+18+19= 70\).

If \(\lfloor x \rfloor = 4, \) the sum of all possible values of \(\lfloor 4x \rfloor\) is \(70\).

If   0 < x < 1 ,     \(\lfloor x \rfloor \lceil x \rceil = 0\cdot1=0\)     but     \(0\nless\sqrt0\)

If   1 < x < 2 ,     \(\lfloor x \rfloor \lceil x \rceil = 1\cdot2=2\)     and    \(1<\sqrt2<2\)

So   \(x=\sqrt2\)   is a solution to  \(\lfloor x \rfloor \lceil x \rceil = x^2\)

If   2 < x < 3 ,     \(\lfloor x \rfloor \lceil x \rceil = 2\cdot3=6\)     and    \(2<\sqrt6<3\)

So   \(x=\sqrt6\)   is a solution to  \(\lfloor x \rfloor \lceil x \rceil = x^2\)

If   3 < x < 4 ,     \(\lfloor x \rfloor \lceil x \rceil = 3\cdot4=12\)     and    \(3<\sqrt{12}<4\)

So   \(x=\sqrt{12}\)   is a solution to  \(\lfloor x \rfloor \lceil x \rceil = x^2\)

The product of the three smallest, positive, non-integer solutions  =  \(\sqrt2\cdot\sqrt6\cdot\sqrt{12}\,=\,\sqrt{144}\,=\,12\)