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Problem 2)

Since BD=AD=AC then triangle ABC is isosceles with AB=AC. Therefore, ∠ABC=∠ACB=x (for some angle measure x ).

Since D is on side BC so BD=AD, then line AD is an angle bisector of ∠A. Therefore, ∠BAD=∠CAD=2∠A​.

Since the angles in a triangle sum to 180∘, then in triangle ABC, ∠A+∠B+∠C=180∘. Substituting x for ∠B and ∠C, we get ∠A+x+x=180∘.

Solving for ∠A, we get ∠A=180∘−2x.

We are also given that ∠BAD=∠CAD=2∠A​. Substituting the expression for ∠A that we just found, we get 2180∘−2x​=∠BAD=∠CAD.

Because the sum of the angles in a triangle is 180∘, then in triangle ABD, ∠BAD+∠ABD+∠ADB=180∘.

Substituting 78∘ for ∠ABD (since BD=AD=AC and triangle ABC is isosceles), we get 2180∘−2x​+78∘+90∘=180∘.

Solving for x, we get x=31∘​.

Therefore, angle ACB = 31 degrees.

Problem 1)

Since ∠ABC=78∘ and ∠ACB=47∘, then ∠BAC=180∘−78∘−47∘=55∘.

Let D be the foot of the altitude from B to AC. Then ∠ABD=∠ABC=78∘.

Triangle ABD is a 30-60-90 triangle, so ∠ADB=60∘ and ∠BAD=30∘.

Since I is the incenter, ∠BAI=∠CAI=∠BAC/2=27.5∘. Then ∠ABI=∠BAI+∠BAD=57.5∘.

Triangle ABI is a 30-57.5-92.5$ triangle, so ∠BID = ∠ABI/2 = 46.25∘​.

The answer is [-2,-4/3]

I'm not sure how to get there, I just took the quiz and this was the correct answer.

Rearrange as

4x^2 - 8x + 4y^2 + 14y  = -12     divide through by 4

x^2  - 2x + y^2 + (7/2)y =  -3      complete the square on x, y

x^2 - 2x + 1  +  y^2 + (7/2)y + 49 /16 =  -3 + 1 + 49/16

(x - 1)^2  + (y + 7/4)^2  = 17 / 16

The radius =   sqrt ( 17/16)  =   sqrt (17) / 4

In order not to have two yellows next to each other,  

the three yellows must occupy squares 1, 3, and 5. 

That leaves squares 2 and 4 to fill, and two colors  

to do it with. There are only four ways. 

Blue Blue     Red Red     Blue Red     Red Blue  

So, the answer is four ways.  

_.

Angle P =  180 - 52  - 79  =   49

Law of Sines

PQ /sin (52°)  = RQ  / sin (49°)

PQ  / sin (52°)  = 5 / sin (49°) 

PQ  =  5 sin (52°) / sin (49° )  ≈  5.22

Area =   (1/2) RQ * PQ sin ( 79°)  =  (1/2) (5) (5.22) sin (79°)  ≈  12.8  yd^2

https://web2.0calc.com/questions/help-geometry_32272

3. An equilateral triangle is rotated about one of its altitudes, sweeping out a cone in space. If the height of the equilateral triangle is 3sqrt3 then find the volume of the cone.

We can imagine (1/2)  of this triangle  being rotated 360°

This will form a  cone  with a height of  3sqrt (3)

The radius = 3sqrt (3) / sqrt (3)  =  3

The volume is    (1/3)pi ( 3)^2 * 3sqrt (3)  = 

9 sqrt (3) pi 

2. A triangle ABC is rotated around side AB sweeping out a solid. If BC = 3, AC = 4  and AB = 5 then find the volume of the solid.

The area of this triangle is  (1/2) (4)(3)  = 6

The height of the triangle is

6 = (1/2) AB * h

12 = 5 * h

h =12/5 

The volume of 1/2 of this solid will be a cone with a radus of 12/5  and a  height of  5/2

So.....the total volume is    2 (1/3) pi ^ r^2 * h   =

(2/3) pi * ( 12/5)^2 (5/2)  =  48pi / 5

1. Volume of the frustum  =   pi * H / 3 * ( r^2 + Rr + R^2)

     Volume of the cylinder  = pi * R^2 * H

And we  know  that

(7/12) pi R^2 * H =   pi * H / 3 * (r^2 +Rr + R^2)       simplify

(7/12) R^2  =  ( r^2 + Rr + R^2) / 3

(21/12) R^2  = r^2 + Rr + R^2

(7/3) R^2 =  r^2 + Rr + (3/3) R^2

(4/3)R^2  = r^2 + Rr              divide through by r^2

(4/3)R^2 / r^2  =  1 + R/r

(4/3) R^2 / r^2 - R/r  - 1  =   0        mulyiply through by 3/4  and  rearrange

R^2/r^2 - (3/4)R/r   =  3/4             let  R/r =  x

x^2  - (3/4)x  =  3/4              complete the square on  x

x^2 - (3/4)x  + 9/64  = 3/4 + 9/64

(x - 3/8)^2  =  57/64        take the positive root

x - (3/8) = sqrt (57) / 8

x =  [ 3 + sqrt (57) ] /  8

R / r  =  [ 3 + sqrt 57 ]  / 8  so......   

r / R   =   8 / [ 3 + sqrt (57) ] =   8 [ 3  -sqrt (57) ] / [ 9 - 57]  =  8 [ sqrt (57) - 3 ] / [ 48 ] = 

[sqrt (57) -3 ] /  6

(a)

9 + 90*2  + 151 * 3  =  642  digits

(b)  Prob of 2   =

No. of 2's

In the ones place

10 in the first hundred

10 in the second hundred

5 in last 50

= 25

In the tens place

10 in the first hundred

10 in the second hundred

10 in the last 50

= 30

In the hundreds place

51  in  the last 51

Total 2's  = [ 25 + 30 + 51  ]  =  106

Prob of a 2  =  106  /  642  =    53  /  321

6 cubes have 1 black face

12 have 2 black faces

8 have 3 black faces

1 has no black faces (in the center)

Probability of  black face  =

(6/27) ( 1/6) + (12/27)(2/6) + ( 8/27) (3/6)  + (1/27) (0/6)   = 1/3

(The first set of parentheses contains the probability of that type of cube is chosen; the second set of parentheses contains the probability of having a black side facing up; mult/add these together for the total probability

Construct a circle centered at the origin with a radius of 1

The equation of this  circle  is

x^2 + y^2   = 1         (1)

Let P, Q lie on this circle

Line  through OP  has the equation

y = 4x

Sub this into (1)  to find the x coordinate of P

x^2 + (4x)^2   =1

17x^2  = 1

x^2 = 1/17

x =1/sqrt 17

y = 4/sqrt 17

P = (1/sqrt 17, 4/sqrt 17)

Similarly, line through OQ  has the equation

y =5x

Sub this into (1) to find the x coordinate of Q

x^2 + (5x)^2  =1

26x^2  = 1

x^2 = 1/26

x =1/sqrt 26

y = 5/sqrt 26

Q = ( 1/sqrt 26,  5/sqrt 26)

Slope of  PQ =   

[ 4/sqrt 17 - 5/sqrt 26 ] / [ 1/sqrt 17 - 1/sqrt 26 ]   =   [ 5 sqrt 26 - 20 ] / [sqrt 26 - 4 ]  =

[ 4sqrt 26 - 5sqrt 17 ] / [ sqrt 26 - sqrt 17 ]  =

[ 4sqrt 26 - 5sqrt 17 ] [ sqrt 26 + sqrt 17 ] / 9  =

[4 * 26  - sqrt (442) - 85 ]  / 9  =

[19 - sqrt 442 ]  /  9   ≈  -0.225

https://web2.0calc.com/questions/coordinates_80366

(p + 2q)  + (6p - 5q) + (14p + 8q)  =  180

21p + 5q  =  180

21p  =  [ 180 - 5q ]

p  =  [ 180 - 5q ] /  21

a^3 + 3ab^2  =  679       (1)

3a^3 - ab^2  =  615     →   9a^3 - 3ab^2  = 1845      (2)

Add (1) , (2)   and we  have

10a^3  =  2524

a^3  =  2524 / 10

a =    (252.4)^(1/3)

Sub this into   (1)

(2524/10) + 3 (252.4)^(1/3) b^2  = 679      solving this  for b produces

b = -4.74353  or b = 4.74353

a - b  = ( 252.4)^(1/3) + 4.74353  ≈   11.06

a - b =  (252.4)^(1/3) - 4.74353 ≈ 1.576

Nevermind, I solved this.

Simplify as

4x^2 + nx + 104 <  0

Set up as an equality

4x^2 + nx + 104 =  0

This will have real solutions when

n^2  - 4 (4) (104) ≥  0

n^2 ≥ 1664

n ≥ 40.79            or   n  ≤ -40.79

So....for our purposes   n < 40.79     and  n > -40.79

The number of  possible values for n  =  [ -40 , 40 ] =   81 possible values

x^2 + y^2   = r^2

(x + 2)^2 + ( y  -1)^2   = r^2            set these equal  for r^2

x^2 + y^2 =  x^2 + 4x +4 + y^2 - 2y + 1

4x - 2y  = -5     (1)

And

x^2 + y^2 = r^2

(x +3)^2 + ( y  -2)^2  = r^2           set these equal for r^2

x^2 + y^2  =  x^2 + 6x + 9 + y^2 - 4y + 4

6x - 4y =  -13      (2)

Multiply (1)  throughby -2  and add to (2)   and we have

-2x = -3

x =3/2

To find y, use (1)

4(3/2)  - 2y = -5

6 - 2y = -5

11 = 2y

y = 11/2

(x, y) = ( 3/2 , 11/ 2)

The vertex is  ( 3,1)  and the points  (0,7) and (6,7) are on the graph

c = 7

1 = a (3)^2  + b(3) + 7

7 = a(6)^2 + b(6) + 7            simplify

9a + 3b =  - 6    →  3a + b = -2 →   b = -2 - 3a     (1)

36a + 6b =  0       (2)

Sub (1)  into (2)

36 a + 6(-2-3a)  = 0

36a - 12 - 18a  =  0

18a =  12

a = 2/3

b = -2  - 3(2/3)  =  -2 - 2 =  -4

a * b * c =    (2/3) * (-4) * 7   =   -56 / 3

The axis of symmetry is 2

If   (4,7)   lies on the graph  then   (2 + 2 , 7)  =  (4, 7)

So

(2 - 2, 7)  =  (0, 7)  also lies  on the  graph

Sorry....see corrections