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We need to split this into two inequalities....we have....

(2/3)t - 1 < t + 7     rearrange as

-8 <  t - (2/3)t

-8 < (1/3)t       multiply both sides by 3

-24 < t

And

t + 7  ≤  -2t + 15     rearrange as

3t ≤ 8     divide through by  3

t ≤ 8/3

So....the solution is

-24 < t ≤  8/3

What is the area of the portion of the circle?

We could use either trig or calculus to solve this.....but a graphical solution seems WAY easier

Look at the graph here : https://www.desmos.com/calculator/cd8js1dkja

The line y = 6 - x   forms a diameter with its intersection of the circle

Notice that the line  y = x - 6 is perpendicular to y = 6- x....so.... it  forms another diameter....and these lines divide the circle into 4 equal parts

So....the part lying to the right of y = 6 - x  and above the x axis =  (1/4) area of circle + (1/8) area of the circle  =

(3/8)  area of the circle

The radius  =  8

So....the area we want is   (3/8) * pi (8)^2  =  24 pi units^2

(a)

x = 2t - 1    ⇔   x + 1 = 2t  ⇒  (x + 1) / 2  = t    (1)

Sub this into the second equation for  t   and we have

y  =  4[ (x + 1) / 2 ] ^2 - 3 [( x + 1 ) / 2)  + 2      simplify

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

y =  x^2 + 2x + 1 - (3/2)x - 3/2  + 2

y = x^2 + (1/2)x  + 3/2    (1)

(b)  x coordinate  of  the vertex  =  [ -1/2]/ [2 * 1]  =  -1/4

y coordinate of the vertex  =  (-1/4)^2 + (1/2)(-1/4) + 3/2  =  23/16

Rearrange (1)  as

y - 3/2  =  x^2 + (1/2)x         complete the square on x

y - 3/2  + 1/16  = x^2 + (1/2)x + 1/16

(y - 23/16)  =  ( x + 1/4)^2

The  vertex  =  ( -1/4, 23/16)

We have that    4a  = 1    so    a  = 1/4

The focus  =   ( 1/4 , 23/16 + 1/4)  = (- 1/4, 27/16)

The directrix  is  y  = 23/16 - 1/4  =   19/16

(c)  When t  = -1    we have the point (-3, 9)

The slope  of the tangent line at any point x  = 2x + 1/2

So...at x = -3, the slope  is  2(-3) + 1/2  =  -11/2

So....the equation of the tangent line  at  (-3, 9)  is

y  = (-11/2) ( x + 3) + 9

y = (-11/2)x - 33/2 + 9

y = (-11/2)x -15/2

See the graph here :  https://www.desmos.com/calculator/ukgeugo1pv

1.10(23+7)=100ℓ+100ℓ

1.1(30) = 200 L

33=200L

33/200 = L           or   .165

I believe that  "theta" is known as the  "argument"

You know the center   -1.5,3       and a point    0,6       work out the equation of the line that connects the two points  (remember y = mx+b ?)  (EDIT: just work out the slope (m) between the two points ( y2-y1)/(x2-x1)

   then m = slope of that line you just calculated....the slope of the normal (perpindicular) line will be  - 1/m   ....then work out the equation of this line given the point 0,6.....    Can you do it?

First one

First term  =  2(1) - 1  = 1

Last term  =2(15) - 1  =  29

Sum  = [ 1st term + last term ] * number of terms  /  2   =  [ 1 + 29] * 29/2  =  15 * 29  = 435

Second one

(A) Number in any row  = 23  - 3(n - 1)   = 26 - 3n

Number in row 6 = top row  =  26 - 3(6)  = 26 - 18  =  8

(B)   Sum =  [  number in 1st row + number in 6th row ] * [ number of rows]  /  2  =

[ 23 + 8 ] * [ 6 ]  / 2  =   31 * 3    =   93  = boxes in whole display

The point is    (2, -2)

r  = √ [ (2)^2 + (-2)^2 ]  = √8  = 2√2

θ  =  arctan ( -2/2)  = arctan (-1)  =   -45°   = -pi/4   =  (2pi - pi/4)  =  (8pi/4 - pi/4)  = 7pi/4

So

2√2 ( cos  7pi/4  +  i sin 7pi/4 )

8(cos 240°   +   i sin 240°  )   =

8 ( -√3/2  - 1/2 i )  =

 - 4√3  -  4 i

( 7 cos pi/2, 7sin pi/2 )  =

(  7 * 0 ,  7 * 1 )  =

(0 , 7)

Yes I can thank you, but can you also teach me how to find the equation of the normal please? 

r  = √[2^2 + 2^2]  =  √8  =  2√2

θ  =  arctan (2/2)   =  arctan (1)  =  45°

So

2√2 ( cos 45°  + i sin 45° )

I'll get ya started....

The equation of a circle in the form   (x-h)^2  + (y-k)^2 = r^2    will give you the center   h, k 

Get the equation to that form

x^2 + 3x       +    y^2 - 6y = 0      'complete the square' for  x   and   y

(x+ 1.5)^2     +   (y -3)^2  = 2.25 + 9

center = -1.5, 3       radius = sqrt  11.25   

Can you keep going from here?

THX, EP!!!!

FYI:    Here is the graph

4x^2 - 2y^2 = 4  ⇒  2x^2 -y^2 = 2   (1)

x^2 + y^2  = 10   (2)

Add (1) and (2)   to get

3x^2 =  12         divide both sides by 3

x^2  = 4           take both roots

x = ±√4

So

x = 2        and     x  = -2

And when x is either of these values, we can find y  as

(2)^2 + y^2  = 10

4 + y^2  = 10

y^2  =  6

y  = ±√6

So.....we have these 4 solutions

(2, √6)  (2, -√6)  ( -2, √6)   (-2, - √6 )

Another option:   re arrange the second equation to this: y^2 = 10 - x^2      Now substitute THIS value of y^2 into the FIRST equation and solve for 'x'.....once you have an 'x' value(s)....sub this value of x into one of the original equations to calculate the corresponding 'y' value(s).....

    let us know what  you find!

A third option would be to graph the two equations and look for the points that they intersect......   try that too !  

I have no idea what you are talking about. 

What table?

You can find your answer here. 

Post the answer so everyone else knows too  :)

Rearrange both equations so that   y^2 is the subjects of the equation.

Then put them together to solve for x

Give that a go and post your working.  Then, if you need it, we can help you more.  :)

Beats me,

Does another mathematician want to give this one a go?

So true, Chris is always within a stone's throw.  Soon he will overtake   

I do not understand the notation.