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OK, RP....you have already done this without realizing it...

Slope of given line = -4        point ( 1, -2)

y = slope ( x - x coordinate) + y coordinate

y = -4 ( x - 1) + (-2)

y = -4x + 4 - 2

y = -4x + 2

NOW this question makes sense......we had it posted yesterday as 140 pillows.

Number of pillows per row =  (2r-1)       r = number of rows

total pillows =     (2r-1)(r) = 120

2r^2-r-120 = 0        Use Quadratic Formula to find   r = 8   ( and -7.5 which we will toss out)

Pillows per row = 2r-1 = 15 pillows/row     8 rows

Square area of the whole framed picture = 32 x 29 = 928 in^2

928-550 = 378 in^2  = area of frame

32 x 29 -( (32-x) (29-x)) = area of frame= 378 in^2    where x = 2 times the width of the frame

928 - x^2 + 61x - 928 = 378

-x^2+61x-378 = 0     Use Quadratic Formula to find x = 54 or 7    (54 is too big...toss it out)

x = 7     width = 1/2 x = 3.5 inches

As a check....does the inside dimension of the frame = the picture size of 550 in^2 ?

(32-7)(29-7) = 550     Check!

sorry it was a typo it's 120 pillows. 

Show that is is true for  n = 1

7 (7¹ - 1)  = 7 (6) = 42      so...true !!!!

Assume it is true for n = k

That is :

7(7^k - 1)   is divisible by 42

Note that we can write this as 

(7* 7^k - 7)  

Prove it is true for n = k + 1

7(7^{k+ 1} - 1)  =

7 ( 7 * 7^k - ( 7 - 6) ) =

7 (  [7 * 7k - 7 ] + 6 ) =

7 ( 7 * 7^k - 7)  + 7*6

Note that the first term is divisible by 42 since ( 7 * 7^k - 7) was assumed to be divisible by 42

And 7*6 is clearly divisible by 42

So ...   7(7^{k+ 1} - 1) is divisible by 42

I assume the spinner and die are independent.  So

\(P[\text{spinner is multiple of 4} \cap \text{dice is factor of 15}] = \\ P[\text{spinner is multiple of 4}]P[\text{dice is factor of 15}] = \\ \dfrac{1}{5} \dfrac{1}{2} = \dfrac{1}{10}\)

\(P_1 : 7(7^1 - 1) = 7\cdot 6 = 42 \text{ is divisible by }42 \text{ so }P_1 \text{ is True}\\ \text{assume }P_n \text{ is True, and prove }P_{n+1} \text{ is True}\\ 7(7^{n+1}-1) = 7(7^n\cdot 7 - 1) = \\ 7(7^n\cdot 7-7+6) = 7\cdot 7(7^n - 1) + 42 \\ \text{we assume }7(7^n -1) \text{ is divisible by 42 i.e. = }42k \text{ for some }k \in \mathbb{N}\\ 7\cdot 7(7^n - 1) + 42 = 7\cdot 42k+42 = 42(7k+1) \\ \text{thus }7(7^{n+1}-1) \text{ is divisible by }42 \text{ and }P_{n+1} \text{ is True. QED}\)

Back in a few....making corrections to my calcs..

\(\sin^2(\theta) + \cos^2(\theta)=1\\ \sin^2(\theta) = 1 - \cos^2(\theta) =\\ 1 - \dfrac{25}{36} = \dfrac{11}{36} \\ \text{so }\sin(\theta) = \sqrt{\dfrac{11}{36}} \text{ or }-\sqrt{\dfrac{11}{36}} \\ \text{ and we must use the quadrant info to determine which one} \\ \text{we are told } \theta \text{ lies in quadrant I, thus }\sin(\theta)\geq 0\\ \text{and it must be that }\sin(\theta) = \sqrt{\dfrac{11}{36}}=\dfrac{11}{6}\\ \text{This is choice C} \)

Here's (2)  using my method

m = (1/2)        point = ( 6, 1)

y = slope ( x -   x coordinate ) + y coordinate

y = (1/2) ( x - 6) + 1    simplfy

y = (1/2)x - 6/2 + 1

y = (1/2)x - 3 + 1

y = (1/2) x - 2 

So....Correct !!!

Correct-o-mundo !

Thanks, EP !!!

Here's another way

(1)   m = 3            point = ( - 1, 4)

Write the equation as

y = slope ( x -   the x coordinate)   + the y coordinate     

So we have

y = 3 ( x -  -1) + 4

y  = 3 ( x + 1) + 4       simplify

y = 3x + 3 + 4

y = 3x + 7

2) y-y1=m(x-x1)

y-1=1/2(x-6)

y-1=1/2x-3

y=1/2x-2

Am I right?

Everytime you add (or subtract ) 360 degrees to an angle, you are right back at your starting point (the original angle)

47 + 360 +360 = 767 degrees

47 - 360 = -313 degrees

OK....you have the correct slope for (5)  =  5/3

So...a perpendicular line will have the slope = -3/5

And an equation for a perpendicular line :   y = (-3/5)x

For (6)

4y - 1 = - 1x     add 1 to both sides

4y  = -1x + 1        divide both sides by 4

y = (-1/4)x + 1/4        the slope is (-1/4)

A perpendicular line has the slope =   4/1  = 4

( "Flip"  -1/4 = -4/1  ..   take the negative of this = - (- 4/1) = 4/1 = 4 )

So....the equation for a perpendicular line is  :    y = 4x

Hope that makes some sense.....!!!

I'll show you how to do the FIRST one....you shoyld be able to do the others....

(y-y1) = m(x-x1)

(y-4) = 3(x-(-1))

y-4 = 3x +3

y=3x+7      ta - da !       Go RP....try the others....

So C ??

Using calculator       arcsin (0.89) = 62.87 degrees (rounded to nearest hundredth)

Hypotenuse of the angle  = sqrt (8^2 + (-7)^2) = sqrt(113)

    sin is opp/hyp = -7/sqrt113

    cos = adj/hyp = 8/sqrt(113)    multiply numerator and denom by sqrt(113) to get    8 * sqrt(113) / 113

In quadrant IV    cos is positive   and sin is negative     

(view cos as 'x'    and sin  as  'y'  to help with these questions)

u_n = 3 * 2^n-1 - 1     

Prove this is true for  n = 1

u₁ = 3 * 2 ^{1 - 1} + 1

u₁ = 3 * 1 - 1 =   2     which is true

Asume that it is true for n = k

That is : 

u_k = 3 * 2^k-1 - 1

Show that it is true for n = k + 1

u_k+1 = 3 * 2 ^{(k + 1 - 1)}  - 1

u_k+1 = 3 * 2^k - 1     (1)

And we were given  that

u_k+1  = 2u_k + 1        (2)

So....by our assumption u_k = 3 * 2^k-1 - 1    (3)

Sub    (3) into (2)

u_k+1  = 2 ( 3 * 2^k-1 - 1) + 1  =

2 * 3 * 2^k-1 - 2 + 1  =

3 * 2 * 2^k-1 - 1 =

3 * ( 2¹ * 2^k-1) - 1 =

3 * ( 2 ^{k - 1 + 1})  - 1   = 

3 * 2^k - 1        which   equals  ( 1)

In quadrant 1     sin and cos are both POSITIVE

cos^2 + sin^2 = 1

(5/6)^2 + sin^2 = 1

sin^2 = 1- 25/36

sin^2 = 11/36

sin = sqrt(11) / 6     (the positive sq rt of   11/36)

Circumference = pi * 2r      The fraction of the total circumference will be   36pi/(pi*2r)

2 pi radians in complete circle

36 pi /(pi *2r)   * 2 pi = 3pi/4 Radians

What program did you use to make that?

1.   -230 degrees = +130 degrees   = quadrant II

2.     Circumference = pi * d

            32pi = 240/360  pi *d

             32 (360/240) = d = 48     r = 1/2 d = 24 units