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For question 1 it looks like a straight line through the points would go through y = 2 when x = 0, so the constant term is 2.

The line would increase by 2 units in y for every 1 unit increase in x, so the gradient is 2.

Hence the best fit line would be y = 2x + 2

Try the others using the same reasoning (if the line decreases as x increases the gradient will be negative).

We have no view of the scatter plots!!

Originally:  C = 4K

After:  C + 20 = 2(K + 20)

Use the first in the second:  

 4K + 20 = 2(K + 20),  or 4K + 20 = 2K + 40 so K = 10

Use this in the first equation.  C = 4*10 or C = 40

Hence Kyle originally had 10 dollars and Christina originally had 40 dollars.

Here's one approach:

Without loss of generality let AB = 1

Let Joe's CDs =J
Let Peggy's CDs=P
J/P =5/3
5P+.5J=.5J + 18, solve for J, P
P = 18 CDs that Peggy started with
J = 30 CDs that Joe started with. So Joe gave Peggy:
30/2 = 15 CDs.

hectictar: You can do this!. Hint: angle AEB=40 degrees, and x=30 degrees !!.

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You write c = square root 653 or - square root 653 for that.

Would i square root 653 or is 653 my final answer?

Distribute -4n to all terms in the bracket

\((n^2-6n+3)(-4n)\\ =-4n^3+24n^2-12n\)

\(c^2 = 22^2+13^2\\ c^2=653\\ c=\pm\sqrt{653}\)

That is 3000000000

(4 + 2i)(3 – 5i) = (3 – 5i)(4 + 2i)

He would have to sell 200 snow cones before he starts making a profit.

Your incorrect Check your math

What is the general form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)?

center at \((x_c=-2,\ y_c= 1 )\)

passing through point at \((x_p=-4,\ y_p= 1 )\)

\(\text{radius}^2 = r^2 = (x_c-x_p)^2+(y_c-y_p)^2\)

general form of the equation of a circle: \((x-x_c)^2 +(y-y_c)^2 = r^2\)

so we have:

\(\small{ \begin{array}{|rcll|} \hline (x-x_c)^2 +(y-y_c)^2 &=& r^2 \qquad | \quad r^2 = (x_c-x_p)^2+(y_c-y_p)^2 \\ (x-x_c)^2 +(y-y_c)^2 &=& (x_c-x_p)^2+(y_c-y_p)^2 \\ x^2-2x_c\cdot x + \not{x_c^2} + y^2-2y_c\cdot y + \not{y_c^2} &=& \not{x_c^2} -2x_cx_p + x_p^2 + \not{y_c^2}-2y_cy_p + y_p^2\\ x^2-2x_c\cdot x + y^2-2y_c\cdot y &=& -2x_cx_p + x_p^2 + -2y_cy_p + y_p^2 \\ x^2-2x_c\cdot x + y^2-2y_c\cdot y + 2x_cx_p - x_p^2 + 2y_cy_p - y_p^2 &=& 0 \\ x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + 2x_cx_p + 2y_cy_p - x_p^2 - y_p^2 &=& 0 \\ x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + x_p\cdot(2x_c-x_p) + y_p\cdot(2y_c-y_p) &=& 0 \\ \hline \end{array} } \)

The general form of the equation of a circle with its center \((x_c,y_c) \)and passing through point \( (x_p,y_p) \) is:

\(\mathbf{x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + x_p\cdot(2x_c-x_p) + y_p\cdot(2y_c-y_p) = 0} \)

\(\small{ \begin{array}{|lrcll|} \hline x_c=-2,\ y_c= 1 \\ x_p=-4,\ y_p= 1 \\\\ & x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + x_p\cdot(2x_c-x_p) + y_p\cdot(2y_c-y_p) &=& 0 \\ & x^2+ y^2 -2\cdot (-2)\cdot x -2\cdot 1\cdot y + (-4)\cdot[2\cdot(-2)-(-4)] + 1\cdot(2\cdot 1-1) &=& 0 \\ & x^2+ y^2 +4x -2y + (-4)\cdot(-4+4) + 1\cdot(2-1) &=& 0 \\ & x^2+ y^2 +4x -2y + 0 + 1\cdot 1 &=& 0 \\ & x^2+ y^2 +4x -2y + 1 &=& 0 \\ \hline \end{array} }\)

The equation of the circle is \(x^2+ y^2 +4x -2y + 1 = 0 \)

You Are correct

\(\dfrac{6}{35}\\ =\dfrac{6}{35}\times 100\%\\ =\dfrac{600}{35}\%\\ =\dfrac{120}{7}\%\)

1st term = 1 = 1/1

2nd term = 1/2 

3rd term = 1/3

...

nth term = 1/n

In this equation, your eventual goal is to isolate y. Let's see how to do this:
 

If you are ever pondering about whether or not this answer is correct, plug the answer you got into your equation and see if the statement is true. :

Therefore, \(y=2\) is the correct and only correct solution for the given equation

9.   domain (-∞, +∞)    the possible x-values

10. range (-∞ , 9]    the possible y values

11. domain [0, 18]   range  [0, 360]

The Ferris wheel takes 12/5 seconds per rotation (a very fast Ferris wheel!) : This is the period.

This is 12/(5*2pi) seconds per radian

The frequency is 5/12 rotations per second (Hz) or 5*2pi/12 radians per second.

.

Are there any requirements or guidelines that the function has to have?

If there aren't any, here's a simple one....

function:      y = (4/5)x         , x ≥ 0

real life situation:     It costs $4.00 to buy 5 apples. What is the price for any number of apples ?

x (input)    =   number of apples

y (output)  =   total price

table:         

\(\frac{\frac1{x^4}*x^8}{1^2}\)                   1^2 = 1

\(=\frac{\frac1{x^4}*x^8}{1}\)              Any number divided by 1 equals the  original number.

\(=\frac1{x^4}*x^8\)           Multiply straight across.

\(=\frac{x^8}{x^4}\)                    Use the exponent rule that says:    \(\frac{x^a}{x^b}=x^{a-b}\)

\(=x^{8-4} \\~\\ =x^4\)