MWizard2k04

Since XY is a diagonal line, let's solve using the Pythagorean theorem of formula c² = a² + b², as we are going to graph a right-angled triangle.

Coordinate input = (x,y)

X to Y

2 - (-8) = 10 steps to the right

7 - (-3) = 10 steps up

If a = 10, and b = 10, then c = XY.

We will approach c, or XY by doing:

c²=a²+b²

c²=10²+10²

c²=200

c=√200

c=√100 √2

c=10√2

c=14.142 units, rounded to 3 decimal places.

Square root of 75 = √25 * 3

                            = √25 √3

                            = 5 √3

Square root of 86 remains √86 as there are no square numbers rooted when factorised.

Quadratic equations consist of ax² + bx = c.

For example,

x² + 2x + 2 = 0

x² + 2x = -2

a + b = c

x = (-b ± √b² - 4ac) / 2a

x = (-2 ± √2² - 4 * 1 * 2) / 2*1

x =  (-2 ± √2² - 8) / 2

x =  (-2 ± √4 - 8) / 2

x = (-2 ± √-4) / 2

x = (-2 ± 2 i) / 2

x = -1 ± i

So this is an example of Quadratic Equation.

Answer:

-9x + 12 / 3

= -9x + 4

Hmm... The question is a bit confusing. I think you mean:

Use the "sub" and "sup" to get base 2 or something raised to any power.

Margot is sewing a ribbon on a seam along the perimeter of a square pillow. The side length of the pillow is 2 x² +1 inches, and its perimeter is 4(2x² + 1) inches. If x = 3.5, how much ribbon does she need?

Pretty obvious. 4(2x² + 1) inches = 4(2*3.5*3.5 + 1)

                                                      = 4(25.5)

                                                      = 102 inches

We 4th graders know that 4 times of one side of a square = perimeter.

You may need to improve this question a bit.

Margot is sewing a ribbon on a seam along the perimeter of a square pillow. The side length of the pillow is 2 x² +1 inches, and its perimeter is 4(2x² + 1) inches. If x = 3.5, how much ribbon does she need?

Extremely easy question; I am a 6th grader, and I know the answer by solving it in 5 seconds.

The answer is 5.

A 5th grader knows that already. I do NOT know your age.

Sorry, I made a calculation error in the first step.

T_n = 3w/10 + 4

T₁ = 3 * 0 / 10 + 4

     = 4

T₂ = 3 * 10 / 10 + 4

    = 7

T_{3 = 3 * 20 / 10 + 4}

    _{= 10}

T₄ = 3 * 30 / 10 + 4

    = 13

T₅ = 13 + 3 = 16

We notice that the pattern increases by 3 when n increases by 1.

So the T chart is:

W        T

0         4

10       7

20       10

30       13

40       16