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In triangle ABC,  the angle bisector of  BAC meets  BC at D.

If   BAC=60°, CAD=45°, and  AD=24, then find the area of triangle ABC.

\(\color{blue}If\ D\ lies\ on\ the\ bisector\ of\ \angle BAC=60°, \angle CAD \ cannot\ be\ equal\ to\ 45°.\)

 !

We can solve this problem using the Pythagorean Theorem applied to both triangles ABC and ACD since we are given a side length and looking for the hypotenuse of each right triangle.

For triangle ABC, we are given that AC=4 and solving for AB (the hypotenuse) using the Pythagorean Theorem: AB2=AC2+BC2=42+x2=42+452, so AB=42+452​.

For triangle ACD, we are given that AC=4 and solving for AD (the hypotenuse) using the Pythagorean Theorem: AD2=AC2+CD2=42+y2=42+632, so AD=42+632​.

To find the difference in length between AB and AD, we subtract the length of AC from each and find the difference: AD−AB=42+632​−42+452​.

We can evaluate these square roots on a calculator and round to the nearest hundredth to find that AD − AB ≈ 8.79 − 6.65 = 2.14 units.

  2x^2 + 3x + 8x - x^2 + 4x + k has a double root    

  combine like terms                          (2x² – x²) + (3x +8x + 4x) + k  

                                                           x² + 11x + k    

  to have a double root,    

  i.e., both roots are the same,   

  the discriminate must equal zero        b² – 4ac = 0    

                                                            (11)² – (4)(1)(k)  =  0    

                                                             121 – 4k  =  0    

                                                                     –4k  =  –121    

                                                                          k  =  121 / 4    

_.

1.   At the vertex of a polygon, the sum of the exterior angle and the interior angle is 180^o.   

      Since the interior angle is 12 times the exterior angle, the exterior angle has to be 15^o.  

      The sum of all exterior angles of a polygon is 360^o, irrespective of the number of sides.  

       Therefore, the polygon has 360 / 15 = 24 vertices.   

       We shall presuppose that a polygon has the same number of sides as it has vertices,    

       Therefore, this polygon has 24 sides.    

_.

                                       y = -2x^2 + 8x - 15 - 3x^2 - 14x + 25    

group like terms              y  =  (–2x² – 3x²) + (8x – 14x) + (–15 + 25)    

combine like terms           y  =  –5x² – 6x + 10    

first derivitive                    ^dy/_dx  =  –10x – 6    

set equal to zero                    0  =  –10x – 6    

                                           10x  =  –6   

                                               x  =  – 0.6    this is the x-coordinate of the vertex

sub x into original                   y  =  [ –5 • (–0.6)² ]  –  [6 • (–0.6) ]  +  10    

                                               y  =  (–5 • 0.36)  –  (–3.6)  +  10    

                                               y  =  –1.8 + 3.6 + 10    

                                               y  =  11.8      this is the y-coordinate of the vertex

the vertex is at                        (–0.6 , 11.8)    

_.

The only constraint for the domain here, is the fact that the inside of the square root also known as the radicand must be greater or equal or 0. 

Essentially, we get \(6-x-x^2-2x^2 \geq 0\)

Then, we simply solve this inequality and there's your domain. Now, I'm not going to write the entire process down for this inequality, so try solving it yourself, and give it a try. It's pretty simple. Combine the like terms to get \(-3x^2 -x + 6 \geq 0\). Plug this into the quadratic formula, and you achieve the asnwer of 

In inequality form, it would be \(-\frac{1+\sqrt{73}}{6} \leq x \leq -\frac{1-\sqrt{73}}{6}\)

In Interval Notation, it would be \([-\frac{1+\sqrt{73}}{6}, -\frac{1-\sqrt{73}}{6}]\)

Thanks!

Find the area of the strip.

Let a be the distance of the base 10 touched by the strip.

Let alpha be the slope angle of the strip.

\(tan(\alpha)=\frac{8}{10-a}\\ sin(\alpha )=\frac{1}{a}\\ \alpha =atan(\frac{8}{10-a})=asin(\frac{1}{a})\\ atan(\frac{8}{10-a})-asin(\frac{1}{a})=0\)

\(\color{green}WolframAlpha\\ \color{green}a=\frac{8\sqrt{163}}{63}-\frac{10}{63}\\\)

\(\color{blue}a=1.4625\\ A=8\cdot 1.4625\\ \color{blue}A=11.7\)

\(\color{blue}The\ area\ of\ the\ strip\ is\ 11.7\ .\)

                                                 x^2-mx+24 = 10   

subtract 10 from both sides      x² – mx + 14  =  0   

integers that multiply to 14        (+7)(+2)   

                                                  (–7)(–2)   

                                                  (+1)(+14)   

                                                  (–1)(–14)  

possible values of m                   –9   

                                                   +9   

                                                  –15   

                                                  +15   

total number of

possible values of m                      4    

_.

                                        5x^2 - 2x + 8 - 2x^2 + 6x + 2 - 2x^2 + 4x - 1   

can be regarded as        (5x² – 2x² – 2x²) + (–2x + 6x + 4x) + (8 + 2 – 1)   

combine like terms            x² + 8x + 9   

discriminant is                   8² – (4)(1)(9)  =  64 – 36  =  28   

The discriminant is the expression under the radical   

in the quadratic formula.   

When a quadratic is stated in the form ax² + bx + c  

the discriminant is b² – 4ac.  

_.

We can divide this problem into two cases:

Case 1: Each row has exactly one child from each family.

Choose one child from each family for the first row: there are 3 choices

.

Arrange the remaining two children in the first row (siblings can't be together): there are 2!=2 ways.

Arrange the second row similarly: 3⋅2=6​ ways total.

Case 2: One row has two children from the same family.

There are two subcases depending on the arrangement of siblings in the rows:

Subcase 2a: The first child in each row is from the same family.

Choose one pair of siblings: 3 choices

.

Arrange the siblings within the pair: 2 ways.

Arrange the remaining 4 children (2 from another pair and 2 from the third pair) in the second row: 4! ways.

Overcount: we've counted the arrangement as if sibling order matters within a pair (which it doesn't) twice (once for each sibling in the first row). So, we divide by 2!⋅2!.

Total: 2!⋅2!3⋅2⋅4!​=36 ways.

Subcase 2b: The first child in the first row is NOT a sibling of the first child in the second row.

Choose one pair to have their children occupy the third seats in each row: 3 choices.

Arrange the remaining 4 children in the first row: 4! ways.

Overcount: similar to subcase 2a, we divide by 2!⋅2! to account for sibling order not mattering.

Total: 2!⋅2!3⋅4!​=36 ways.

Since Cases 1 and 2 are mutually exclusive, to find the total number of arrangements, we simply add the number of arrangements from each case:

6 + 36 + 36 = 78​.

This gives us a final answer of 78.

We can solve this problem by considering the empty spaces between the circles. There are 11 spaces between the circles, since there are 12 circles forming a circle.

Since no two chosen circles can be next to each other, the chosen circles must correspond to every other empty space. There are C(11,3) = 165​ ways to choose 3 empty spaces out of 11.

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