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Second one: 

x=y

y =x^2+3x-3

Substitute  x = y into the second equation

y = y^2 +3y-3   

0 = y^2 +2y -3   factor

0 = (y+3)(y-1)                    so y = -3  or 1      and since x= y 

                                                -3,-3   or  1,1   are solutions 

What is the volume of this square pyramid?

Enter your answer in the box.

 ft³

This figure is made up of a quadrilateral and a semicircle.

What is the area of this figure?

Use 3.14 for pi. Round your final answer to the nearest tenth.

15.7 units²

17.9 units²

25.7 units²

31.4 units²

Which postulate or theorem can be used to prove that△BCD is similar to △EFG?

ASA Similarity Theorem ​

AA Similarity Postulate

SAS Similarity Theorem

SSS Similarity Theorem

4a^5          2b^2

____   *   _____             swap denominators

7b^4          2a^4

4a^5         2b^2

____   *   ______

2 a^4        7b^4

2a  *    2

         ______

          7 b^2

4a

___

7b^4

Second answer, NSS!!!

x^2  - 16                x^2 + 5x + 4

__________    ÷    ___________

x^2 + 5x + 6          x^2 - 2x  - 8

(x + 4) (x - 4)             (x + 1) ( x + 4)

____________    ÷    ___________

(x + 3) ( x + 2)            (x - 4) ( x + 2)        invert the second faction and multiply

(x + 4) ( x - 4) (x - 4) ( x + 2)

______________________

(x + 3) ( x + 2) (x + 1) (x + 4)

(x - 4)^2 

____________       x cannot be    -3,  -2, 4, -1, -4

(x + 3) ( x + 1)

Third answer

Let's do it this way.....let  x= n^2

Substitute:

x^2 -11x+30 / x^2-7x+10      factor top and bottom

(x-6)(x-5)  / (x-2)(x-5)           Now put the n's back in

(n^-6)(n^2-5) / (n^2-2)(n^2-5)      Now you can see that n^2 cannot equal 2 or 5

                                                            so n cannot equal +-sqrt2  or +- sqrt 5

cancel out the (n^2-5) in num and denom

(n^2-6) / (n^2-2)

Welcome Arriane!

YAH!!!!!!!

We can use the Secant-Tangent Theorem first to find BC

Specifically

The length of the tangent squared  =  the product of length of the tangent * its exterior segment

8^2  =  (4 + BC) * 4  simplify

64  =  16 + 4BC         subtract  16 from both sides

48  =  4BC    divide both sides by 4

12  =  BC

So......AD  = 8  and  AC =   4 + BC  =  16

And these will form two legs of a right triangle with CD as a hypotenuse

So

CD   =  √  ( 8^2 + 16^2)  = √ [ 8^2 ( 1 + 2^2) ]  =   8√5   units

I actually thought the first one was D

Thanks for your help!

First one:

The solutions to the system are the points where the two lines intersect. Since they don't intersect at any points, there is no solution.

Second one:

F - 32  =  \(\frac95\)(k - 273.15)                    We want to solve this equation for  k .

                                                         Multiply both sides of the equation by  \(\frac59\)

\(\frac59\) * (F - 32)  =  \(\frac59\) * \(\frac95\)(k - 273.15)

\(\frac59\)(F - 32)  =  k - 273.15

                                                         Add  273.15  to both sides of the equation.

\(\frac59\)(F - 32) + 273.15  =  k

k  =  \(\frac59\)(F - 32) + 273.15

"problem" with the "Waiting for Moderation" messages has been fixed

This site is for if you need help only. How the heck am I a nerd?

Who said I was giving up on math? There are just certain questions that I don't understand or just can't answer. And btw, I do online school, they don't give us books they only give us the material and videos that don't even go with the material.

U WOT M8?

3xy - 4x^2 - 36y + 48x = 0   (1)

x^2 - 2y^2 = 16   (2)

Look at the graph here :

https://www.desmos.com/calculator/kfu7s3izs1

Note that, oddly, the first graph is a straight line ......this is known as a "degenerate conic" which never intersects the second function (a hyperbola)....so....no solutions

Thanks so much 

I  =  6.5%  =  .065

Yearly

A  =  1234 ( 1 + .065)^4   ≈ $1587.50

Monthly

A  =  1234 ( 1 + .065/12)^(4*12)  ≈ $1599.29

Continuously   ....this uses   Pe^(r * t)

So  we have

1234 *e^(.065 * 4)   ≈ $1600.41

Note that continous compounding isn't much more than monthly  !!!

The common ratio between terms is  1.1

If we are just trying to find the  nth term, we can use

a_n  =  (2.2)(1.1)^(n - 1)

If you are summing these terms we have...

        inf

2.2 * ∑  (1.1)^(n - 1)  

       n=1

Year 2      28.4

Year 4      29.2      29.2-28.4 = .8

the winning time increased by  .8 sec/2 yr = .4 sec per year

A No

B  No

C No

d  Yes