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I don't think your image is showing up.

What you have to do is make a graph of with the number of animals along the bottom and cost up the side.

there will be two straight line graphs. One for each scenario.  There will be a cross over point where they are both the same cost (probably a part of an animal)  But, for any given number of animals, the graph on the top will always be the dearer option. 

Now give it a go.  :)

Please no one answer is full.  

thank you for your help! (: the explanation has helped

By vieta's formulas, given a cube(it can be any polynomial) ax³+bx²+cx+d = 0

and roots r,s, and t:

r+s+t = -b/a

rs +rt + st = c/a

rst = d/a

you could try solving for these values in terms of the ones that you want(for example, with 2r, 2s and 2t, rewriting these equations using those respective terms). I'm sure that this would be one approach that would work. 

If you want more information to learn more about vieta's, the internet is always a great tool to search up information regarding it.

\((x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2\)

The problem gives us that:

\(x^2-y^2 =5\)

and 

\(xy =6\)

Realize that we already have all the necessary values to determine \(x^2+y^2 \) without individually solving for x or y

Substituting the values of \(x^2-y^2\) and \(xy\) that we already have:

\((x^2+y^2)^2 = (5)^2 + (2*6)^2\)

\((x^2+y^2)^2=25+(12)^2= 25 + 144 = 169\)

Now we square root both sides to get:

\(x^2+y^2 = \pm13\). The problem doesn't directly limit x and y to real numbers, so we could have an answer of -13(square rooting 169 gives a negative and positive solution). The implied answer however would be 13.

If we cube both sides of this equation(I couldn't see an easy way out of this), we get:

\(343x = x^3 + 3x^2 * 6 + 3x * 6^2 + 6^3 \)

\(343x = x^3 + 18x^2 + 108x + 216\)

Subtracting 343x on both sides, we get:

\(x^3 + 18x^2 -235x + 216 = 0\)

Realize that the sum of the coefficients is 0, which means that 1 is a root of this polynomial. Then we can factor out x-1 from this.

Doing polynomial division, we get this factored into:

\((x-1)(x^2+19x-216)\)

We factor the right hand quadratic into:

\(x^2+19x-216 = (x-8)(x+27)\)

We have then factored our cubic into:

\((x-1)(x-8)(x+27) = 0\)

This gives us our three solutions(by the fundamental theorem of algebra, there should be 3), as :
 

1, 8, and -27

Could you make the question clearer? Are we supposed to be finding when the options are equal to each other? Then I would be happy to help! :)

We can make use of the power of a point theorem, a useful application for geometry. 

It has 3 variations, and in this case, I'll use 1 of them : the case with two secants intersecting a circle(in other words, the lines intersect the circle at two points each).

It states:

Given the two secants in this problem GJ and GL,

GH * GJ = GK * GL

we are given that:

GH = 13

GJ = z + 13

GK = 9

GL = 39

Substituting, we get:

13(z+13) = 9 * 39

Dividing by 13 on both sides, we get:

z+ 13 = 9 * 3

z + 13 = 27

z = 14

secant = 1/cos

Dividing by -sqrt3 on both sides

\(\sec{\theta} = -\frac2{\sqrt3}\)

Rationalizing the denominator by multiplying it by sqrt3/sqrt3(which is one):

\(\sec{\theta} = -2\sqrt{3}/3\)

Substituting in 1/cos:

\(1/\cos{\theta} = -2\sqrt{3}/3\)

\(\cos{\theta} = -3/2\sqrt{3}\)

Rationalizing like before by multiplying by sqrt3 / sqrt 3, we get:

\(\cos{\theta} = -3\sqrt{3}/6 = -\sqrt3/2\)

The values for which this is true between [0, 2pi] are :

\(\cos{5\pi/6}\) radians (cos 150 degrees) and \(\cos{7\pi/6} \)(cos 210 degrees). The general forms for these are then indeed your answer because the period of cosine is 2pi.

Did you learn anything that time qwerty?

I see you are so grateful when people try to help you that much of the time you cannot even be bothered responding.

Draw it using Desmos

What happened to the question?

I assume it was edited out but there is no edit message ??

I have seen this a bit lately.

Thanks guest,

Yes I know that is what he means but I have to rewrite the question properly before I can think it through.

I just cannot look at those dollar signs and ignore them, I have to physically remove them before I start on the problem.

I'm just not prepared to do that.

If they are too lazy to post their questions the best possible, then I expect they will be far too lazy to learn from any response.

If you want to find the horizontal side that you don't know, subtract 10 in from 17 in to get 7 in.

If you want to find the vertical side that you don't know, subtract 11 in from19 in to get 8 in.

You now know all the sides:  add 10 in to 19 in to 17 in to 11 in to 7 in to 8 in to get your answer.

Definitely cannot divide by 0.     LOL

dy/dx for e^x + y = y

Perhaps I do not understand the question. Maybe more than just perhaps  

e^x + y = y

e^x=0

Never happens so you cannot find a tangent.   ???

e^x and also its gradient do approach 0 as x approaches -infinity though. 

Maybe another mathematician should look at this. Maybe my definitions are not quite right.

the question does not make sense to me.

To find out how many cookies (at 35 cents each) must be sold to make more than $15.00,

divide $15.00 by 0.35  --->   42.85... cookies

So she must sell at least 43 cookies.

Checking:  43 x 0.35  =  15.05 ...

Continue line EF to the left so that it intersects line BCD at point X

Since FEX is parallel to AB, angle(CXE)  =  180^o - angle(ABC)  =  180^o - 120^o  =  60^o.

In triangle(CXE), angle(CXE) = 60^o and angle(ECD) = 55^o, so angle(XEC) = 65^o.

This means that angle(CEF) = 115^o.

Thank you both so much, it helped a lot!  Hope you have a great day! :)

Let X be the point where the external tangent meets circle A.

Let Y be the point where the external tangent meets circle B.

Because AX is perpendicular to CYX and BY is perpendicular to CYX, triangle(ACX) is similar to triangle(BCY).

Therefore:  CB : CA  =  BY : AX.

Since  CA  =  CB + BA  =  CB + 17  and  BY  =  7  and  AX  =  10:

--->   CB : CB + 17  =  7 : 10

--->  10 · CB  =  7 · (CB + 17)

Can you finish it from here?

A circle centered at A with radius 10 is externally tangent to a circle centered at B with radius 7.  A line that is externally tangent to both circles is drawn, where both circles lie on the same side of the line. This line intersects line AB at C. Find the length BC.

Which equation can be simplified to find the inverse of y = 2x2? 

and its just the power of 2 not 4

I am sorry, but you are not being very clear here. I assume you mean \(y = (2x)^2\), or \(y = 4x^2\). However, please specify what you are asking about.

48 

count it

Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,1).

.