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cos (theta)  =  2 / 3    =   x   / r

We need to find  y   =  sqrt  ( r^2 - x^2)  = sqrt ( 3^2 - 2^2)  = sqrt ( 9 - 4)  = sqrt (5)

sin (theta)  =  y  / r  =   sqrt (5)  / 3

tan ( theta) =  y  / x  =     sqrt (5)  / 2

1) The same question was posted here: https://web2.0calc.com/questions/help_55674

2) 

Let x and y represent the numbers of Platter A and Platter B the organizer needs to purchase.. 

  x + 2y ≥ 80.  80+ hamburgers are required

  3x + y ≥ 95 . 95+hot dogs are required

  5x +8y ≥ 380.38+ chicken wings are required

  16x +20y = c 

The vertex is (x, y) = (20, 35). The other vertex close to the origin is (60, 10). The cost of that order would be $1310.

The value of an order of 20 Platter A and 35 Platter B is $1020

that's not an excuse :))

(a - 1) ( a + 1) ( a + 2)  - (a - 2) (a + 1)  =

(a + 1)  [  (a - 1) ( a + 2)  - (a - 2)]   =

(a + 1)  [ [ a^2 + a - 2  - a + 2 ]  =

(a + 1) [ a^2 ] =

a^3 + a^2

then chill w that. pls.

x + y   = 7/13

x - y   =  1/91

Note that   x^2 - y^2  can be factored as  ( x + y) (x - y)  .....substituting....we have  that

x^2 - y^2  =  ( 7/13) ( 1/91)   =   (7/91) (1/13)  =  (1/13)(1/13)  =  1 / 169

I've just answered your question about the S and T intersection thingy. Take a look at it! I've also given you hints on a lot of your problems.

S=9-2t

S=9-2(3s+1)

S=9-6s-2

7S=7

S=1

(Use the Substitution Property)

t=3S+1

t=3+1

t=4

Therefore, the answer is (1,4)

Hope this helped!

x^2  + y^2  =  4x  +  4y      rearrange  as

x^2 - 4x  + y^2  - 4y  =   0        complete the square  on x and y

x^2 - 4x + 4  + y^2 - 4y + 4   =  4 + 4      simplify

(x - 2)^2  + (y - 2)^2  =  8

x  will be maximzed when  y  = 2   ....so we  have that

(x - 2)^2  + (2 - 2)^2  =  8

(x - 2)^2  =  8       take the positive root

x  - 2  = √8

x = √8 + 2

x  = 2√2 + 2

x = 2 (√2 + 1)

im still a bit confused

Thanks!!

Here's a hint: Since you know that QSR is 90 degrees, SQR would be 25 degrees. Also, using this information, (Plus a little more) you can find out P.

Hope this helped!

Sorry...I read 3x^2  as 3x^3....see my corrected answer  !!!

I will give you the steps but try to do the problem by yourself.  

Start by finding the volume of the tank

Multiply 1.75 ft and 1.25 ft then covert the sum to inches by multiplying it by 12.

Then multiply that amount by the ll inches and you have the volume.

Since the tank is 20% full divide the volume by five, take the number you get from that and subtract it from the origanl volume.

Now you have the amount you need to fill.

Since it is 3$ a gallon and 231 cubed inches is a gallon 77 cubed inches is 1$.

Divide the amount you need to fill by 77 and you get the cost to fill the tank.

i was very stressed my apologies

This same question was already posted here: https://web2.0calc.com/questions/please-help_19950

Please look at this, instead of repeating a post.

Hope this helped! 

This one is a little difficult

Call Pierre's height above the ground, H

Call the  distance from the base of the bldg to the sports car, D

So...we have this relationship

tan 21°  = H / D    ⇒     D tan 21°  =  H     (1)

Next.....call the distance from the  base of the bldg to the  dump truck = D + 600

So  we have this

tan 14°  =   H  / (D + 600)       sub (1)  in for H

tan14°   =  ( D tan21° )  / (D + 600)       rearrange as

D tan 14° + 600 tan 14°  = D tan 21°

600 tan 14°  =   D ( tan 21° - tan 14°)

[ 600 tan 14° ]  / [ tan 21° - tan 14°]  =  D  = 1112 ft

Sub this back into (1)  to find H  = 1112 tan 21° ≈  426.86 ft

GOOD JOB, JENNY   !!!!!

I don't think this is necessary, considering that me and a lot of others are looking at your questions right now. Be patient! 

999,999,999,996,000,000,000,004=11 zeros

I do appreciate the help but you got the problem wrong.  Could you try again?

Note  that if we drew a perpedicular to both  l and k   we would have an (irregular) pentagon

Call the intersection points of this perpendicular  to l and  k , D and E, respectively

And since  l and k  are parallel then angle   ADE   = angle  BED   =  90°

And since the sum of the interior angles of a pentagon  = 540°

So

angle A + angle C + angle B + angle D + angle E  = 540

120   +  80  +  angle B + 90   +  90    = 540    simplify

380 + angle B =  540   subtract  380  from both sides

angle B  = 160°

1)    That angle is  105º        ( Tthis works with 30-60-90 triangles only )

I tried using  20-70-90 triangle, and the result was 108º, plus black (red) line doesn't bisect the top right angle.

9x^3  -3x  + 1  + h(x)  =  3x^2  - 5x + 3

h(x)  =  [ 3x^2 - 5x + 3  ] -  [ 9x^3  - 3x + 1 ]

h(x)  = -9x^3 + 3x^2 - 2x + 2

CORRECTED  !!!!

This  leads to an interesting result .....  [ the shaded areas are known as "lunes" ]

The area of the triangle is  (1/2)(BC)(BA)

The area  of the whole semi-circle  is   (1/2)pi * (AC/2)^2  =  (pi/8)AC^2 =  (pi/8) (BC^2 + AC^2)

So....the sum of the areas  between the semi-circle  and the triangle =   (pi/8)AC^2  - (1/2)(BC)(BA)

The area  of  the semi-circle  constructed on BC  = (1/2)(pi)(BC/2)^2 = (pi/8)(BC)^2

The area  of the semi-circle  constructed on AC  is similar =  (pi/8)(AC)^2

So......the  sum of the shaded areas  =

Area of  the two-semi-circles  - area  between the semi-circle and  the triangle  =

(pi/8) (BC^2 + BA^2) - [ (pi/8)AC^2 - (1/2)(BC)(BA) ]  =

(pi/8) ( AC)^2  - ( pi/8)(AC)^2  + (1/2)(BC)(BA)  =

(1/2)(BC)(BA)    =   the area of the right triangle