All Answers 352741 Answers

In a geometric sequence, the ratio between successive terms is constant. In this case, the common ratio is -1/3, which means each term is -1/3 times the term before it.

For the sequence to be greater than 1, the terms need to be positive. Since the first term (1000) is positive and the common ratio is negative, the sequence will alternate between positive and negative terms.

Since there are 400 terms, half (200) will be positive and the other half negative. However, the question asks for terms greater than 1, not just positive terms.

Finding the first term greater than 1:

We can analyze the sequence using the formula for the nth term of a geometric sequence:

tn = a * r^(n-1)

tn: nth term

a: first term (1000)

r: common ratio (-1/3)

n: term number

We want to find the value of n (n = k) where tk > 1.

Since the terms alternate in sign, we need to consider the even terms (positive terms). Let's rewrite the formula for even terms (n = 2k):

t(2k) = 1000 * (-1/3)^(2k-1)

For tk to be greater than 1, we need the exponent of -1/3 to be negative enough to make the result positive.

The smallest even term that satisfies this condition is t4 (fourth term), where k = 2.

t4 = 1000 * (-1/3)^(2(2)-1) = 1000 * (-1/9) = -100/9 < 1

t6 (sixth term) = 1000 * (-1/3)^(2(3)-1) = 1000 * (1/27) = 100/27 > 1

Therefore, the first term greater than 1 is the sixth term (n = 6).

Counting terms greater than 1:

Since the sequence alternates in sign, every two terms after the sixth term will have one term greater than 1. With 400 terms, there are 200 even terms (positive or negative).

Out of these 200 terms, the first 4 (t2, t4, t6, and t8) won't be greater than 1. The remaining 196 terms will have 98 terms greater than 1 (every other term).

So, there are 98 terms greater than 1 in the sequence.

Something like this  

Law of Cosines   {   cos (angle BDA)  =  -cos (angle CDA)  }

BA^2   = BD^2  + DA^2   -2 (BD * DA) cos (BDA)

AC^2   = DC^2  + DA^2  - 2 (BD * DA) (CDA)

(8 + r)^2  = 2^2  + (10 -r)^2  -2 [ 2 (10-r) ] (-cos (CDA))

(2 + r^2)  = 8^2  + (10 - r)^2 - 2 [ 8 (10-r)] (cos (CDA))

(8 + r)^2  = 2^2  + (10 -r)^2  +2 [ 2 (10-r) ] (cos (CDA))

(2 + r^2)  = 8^2  + (10 - r)^2 - 2 [ 8 (10-r)] (cos (CDA))

Equating  Cosines

[(8 + r)^2  - 2^2 -(10 - r)^2] / [ 4 (10 -r))] = [ (2 + r)^2  - 8^2 - (10-r)^2] / [ -16 (10 -r) ]

[ r^2 + 16r + 64 - 4 - r^2 + 20r -100 ] / 4  =  [ r^2 + 4r + 4 - 64 - r^2 + 20r - 100] / -16

[ 36r -40 ] / 4 =  [ 24r -160 ] / -16

- 4 [ 36r -40 ] =  24r - 160

-144r  + 160  = 24r - 160

24r + 144r  =  320

r [ 168 ] = 320

r = 320 / 168   =  40/21  ≈ 1.9

To solve the equation, we first try to get rid of the denominators.

Multiplying both sides of the equation by (x+1)(x+2), we get [x^2 + 2x + x^2 + x = kx(x + 1)(x + 2).]

Expanding the right side, we get [2x^2 + 3x = kx^3 + kx^2 + 2kx.]

Moving all the terms to one side, we get [0 = kx^3 + (k - 1)x^2 + (2k - 3)x.]

Since we are looking for complex solutions, we can treat x as a complex number.

For the equation to have exactly two complex solutions, the discriminant of the quadratic part must be nonzero.

That is, [(k - 1)^2 - 4 \cdot k \cdot (2k - 3) > 0.]

Expanding, we get [k^2 - 2k + 1 - 8k^2 + 24k - 12 > 0.]

This simplifies to [-7k^2 + 22k - 11 > 0.]

We can factor this as [(k - 1)(7k - 11) > 0.]

Thus, k must be in one of the intervals (−∞,1/7)∪(11/7,∞).

Now, let's check that for these values of k, the equation has exactly two complex solutions.

For k=0, the equation becomes x+1x​+x+2x​=0, which has the complex solutions x=−1 and x=−2

For k=2, the equation becomes x+1x​+x+2x​=2x, which has the complex solutions x=0 and x=−23​.

Thus, the ony possible k is 11/7.

          A

                       D

                 4

          E            8           C       

 triangle  ADE  similar to triangle  AEC

AE / DE  = AC / EC

h / 4  = sqrt [h^2 + 8^2] / 8  

8h  =  4sqrt [ h^2 + 64]

2h  = sqrt [ h^2 + 64 ]       square both sides

4h^2  = h^2  +  64

3h^2    =  64

h^2  =  64 / 3

h = 8 / sqrt 3  ≈   4.6

                              C

            B                             D

                           H

                          144

A  66                                                 B

This is impossible  ....the sum  of the angles in  triangle AHB  > 180°

Note that  triangle ADO  is similar to triangle AEC

AO / DO  =  AC / EC

AO  = 3 - r

DO  = r

AC =  sqrt  [ 3^2 + 3^2 ] =  3sqrt [2]

EC  =  3

(3 - r)  / r =  3sqrt (2 )  / 3

  3 - r  = r *sqrt (2)

3 = r [ 1 + sqrt 2]

r = 3 / [1 + sqrt 2] ≈ 1.24

Volume of sphere   = (4/3)(pi)(1.24)^3  ≈  2.54 pi  =  A

Volume of cone = (1/3) pi (3^2) * 3  = 9pi  =  B

A / B  ≈  2.54 / 9  =   127 / 450

Number of  diagonals  of a regular polygon  = n ( n - 3)  / 2     where n is the number of sides

So

(n) ( n - 3) / 2 =  2n

n^2 - 3n  =  4n

n^2 - 7n =  0

(n) ( n - 7)  = 0

n - 7  = 0 

n =  7

Area of a hexagon  =  6 (1/2) s^2  sin (60°)    where s is the side length

3 =  6 (1/2) s^2 *sqrt (3)  / 2

3 = 3 s^2 * sqrt (3)  / 2

s^2  = 2 / sqrt (3)

s = sqrt  [ 2  / sqrt (3) ]   ≈ 1.075 

Using the square of the distances to equate R^2

(x - 22)^2 + (y - 15)^2  = (x -22)^2  + ( y + 29)^2

y^2 - 30y + 225  =  y^2 + 58y + 841

-30y - 58y  =  841  - 225

-88y = 616

y = -616 / 88  = -7

To find x

(x -22)^2 + (-7-15)^2  = (x + 25)^2 + ( -7 -0)^2

x^2 - 44x + 484 + 484  = x^2 + 50x + 625 + 49

-44x + 968 = 50x + 674

968 - 674 =  50x + 44x

294 = 94x

x = 294/94 =  147 / 47

Center  = ( 147/47 , -7)

https://www.desmos.com/calculator/ecmqlf6gaz

3.  Let  QR   =   8 + x

Let M  be the point of tangency with the radius of the large circle  and N be the same  with the smaller circle

We have similar triangles    MPR    and NQR

PR / MP =   QR / QN

(28 + x)  / 12  =  (8 + x) /  8

8(28+x)  = 12 ( 8 + x)

224 + 8x  = 96 + 12x

224  - 96 =   4x

128  = 4x

128/4  = x  =  32

QR =  8 + 32  =  40

5D =   the man hours for  the first job where D is the number of days

6(D  - 4)   = the man hrs if  one additional person is hired

Since the man hours are equal

5D = 6D - 24

D  = 24

5D  = 5 (24)  =   120 man hours

So ....let N be the number of  workers needed to complete the job 20 days earlier

N ( 24 - 20)   = 120

4N  = 120

N =  30

30  - 5  =  25   additional workers should be hired 

The smaller  gear  completes  72 revs per minute

This  is     72 * 2pi * 3  =  432pi  cm / min

They both will have the same linear velocity   

They will have different angular velocities

Exactly correct, hairyberry  .....

The shaded area always  = 25 pi   for  any R, r   such that

R^2  - r^2  =  5^2

See my response below

2.  This  problem  has no unique solution for R , r

But it does have a definite solution for the shaded area

R = radius of  large circle

r = radius of  small circle

R^2  - r^2   = (chord length / 2)^2

R^2 - r^2  =  5^2

One obvious  integer solution is  that R  = 13  and r =  12  since R^2 - r^2 =  5^2

Shaded area =  pi ( 13^2 - 12^2)  =  pi [ 169 -144 ]  =  25 pi

But

R^2  - 5^2  =  r^2

If we let  R = 20

And r  =  sqrt (375)

The shaded area =  pi (20^2 - (sqrt 375)^2 )  =  pi [ 400 - 375] =  25 pi

In general, any  R^2 - r^2   producing 5^2 will work

Thanks booboo44, I realized it is possible , thanks to your work.

But I'm getting a different answer, is there anything wrong with my work?

Set the radius of the outer circle as y.

Set the radius of the inner circle as x.

We want:

\(y^2\pi-x^2\pi=\pi(y^2-x^2)\).

By the pythagorean theorem, we know

\(y^2-x^2=5^2=25\).

Therefore the area is \(25\pi\)

Thank you so much!

Use the formula for area of a regular hexagon.

You get \(27\sqrt3 \over 2\).

Rounded to the nearest thousandth, the answer is 28.383

For problem 2:

Let's denote the following:

O: Center of both circles

A and B: Endpoints of the chord (chord length AB = 10)

P: Point of tangency between the chord and the smaller circle

r: Radius of the smaller circle

R: Radius of the larger circle

Since the chord is tangent to the smaller circle at point P, we know that OP is perpendicular to AB. Additionally, since O is the center of both circles, line segment OP bisects chord AB (AP = BP = 5).

Finding the Radius of the Smaller Circle (r):

Using the Pythagorean Theorem in right triangle APO:

AP^2 + OP^2 = r^2 (since AP = BP = 5)

5^2 + OP^2 = r^2

We don't have the value of OP yet, but we can find it using the information about the larger circle and the chord.

Finding the Radius of the Larger Circle (R):

In right triangle OBP:

OB^2 + OP^2 = R^2 (since OB = half the chord length = 5)

5^2 + OP^2 = R^2

Relating the Radii (R and r):

Since the chord is tangent to the smaller circle, the distance between the center (O) and the point of tangency (P) is the radius of the smaller circle (r).

Additionally, this distance (OP) is also the difference between the radii of the larger and smaller circles (R - r).

Therefore, we have: OP = r and OP = R - r

Setting these two expressions equal to each other:

r = R - r

Solving for r (the smaller radius):

2r = R r = R/2

Substitute r in terms of R back into Equation 1:

5^2 + (R/2)^2 = R^2

Expand and rearrange:

R^2 - 5R - 25 = 0

Factor the equation:

(R - 10)(R + 2.5) = 0

Since the radius cannot be negative, we have R = 10.

Therefore, the radius of the smaller circle (r) is r = R/2 = 5.

Area of the Ring-Shaped Region:

The area of the ring is the difference between the area of the larger circle and the area of the smaller circle:

Area of Ring = π * R^2 - π * r^2

Substitute the values of R and r:

Area of Ring = π * (10)^2 - π * (5)^2

Area of Ring = π * (100 - 25)

Area of Ring = 75π

Therefore, the area of the ring-shaped region is 75π square units.

Note: I will use [] to denote area.

[ABCD] = AD * DC

[ADU] = [BCT] = AD*DU/2 = (AD*DC/3)/2=AD*DC/6

[ADU]+[BCT]=AD*DC/3

[ABTU] = [ABCD]-([ADU]+[BCT]) = 2/3*(AD*DC) = 120.

The largest term in the binomial expansion of (a+b)n occurs when k = n/2 (for even n) or k = (n - 1)/2 (for odd n) in the binomial coefficient \binom{n}{k} = \frac{n!}{k!(n-k)!}.

In this case, n = 31 (odd). Therefore, the largest term occurs when k = (31 - 1)/2 = 15.

The term we want is \binom{31}{15} = \frac{31!}{15!16!}. We can re-write this as:

\binom{31}{15} = \frac{31 \times 30 \times 29 \times \cdots \times 16}{16 \times 15 \times 14 \times \cdots \times 1}

Notice that most of the terms in the numerator and denominator cancel out, leaving us with:

\binom{31}{15} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1} = 15450

The denominator, b, is thus 3 \times 2 \times 1 = \boxed{6}.

2)

Is this question complete? I think it might be non-unique.

4)

Because PT and PU are both tangents, PT = PU (this is easy to prove using HL congruency), and PTO is congruent to PUO.

Because PTO = 90, then POT = 180-90-33=57. These two triangles are congruent. 

So arc TU has a length 57*2=114.