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So, we have:

$60 + a$, $120 + a$, and $140 + a$

For a three-term geometric sequence, the ratio between consecutive terms is constant.

Thus, we can write:

$\frac{120 + a}{60 + a} = \frac{140 + a}{120 + a}$

Cross multiplying:

$(120 + a)^2 = (60 + a)(140 + a)$

Expanding both sides:

$14400 + 240a + a^2 = 8400 + 200a + 140a + a^2$

$14400 + 240a + a^2 = 8400 + 340a + a^2$

$6000 = 100a$

$a = 60$

Therefore, the common ratio of the resulting sequence is $60$.

Simplify the inequality:

−8≤X+17<−3−5X

Subtract 17 from all parts of the inequality:

−8−17≤X+17−17<−3−5X−17

−25≤X<−20−5X

Now, isolate X on one side of the inequality:

Add 5X to all parts of the inequality:

5X−25≤X+5X<−20−5X+5X

5X−25≤6X<−20

Subtract 5X from all parts of the inequality:

5X−25−5X≤6X−5X<−20−5X

−25≤X<−20

So, the values of X that satisfy the inequality are in the interval [−25,−20)

16 * r^3 =  26

r^3  = 26/16

r^3  = 13/8

r = (13/8)^(1/3)

24th term =  16 * (13/8)^(1/3) ≈ 18.81

Another way

8(a + 2) = b (a + 1)

8a + 16 - ab - b  =  0

8a + 16 - ab - b + (-1 * 8)  =  (-1*8)

8a + 8  - ab -b  =  -8

8 ( a + 1)  - b ( a + 1)  =  -8

(a + 1) ( 8 - b)   =  -8

Factors  of -8             a        b

-2  4                          -3       4

4  -2                           3      10

-4  2                          -5      6

2  -4                           1     12

1  -8                           0     16

-8  1                          -9      7

-1  8                          -2      0

8  -1                          7       9

https://web2.0calc.com/questions/sum_93

r =  24 / 16 =  3/2

28th term  =  24 (3/2)^4  =  24 * (81 / 16)  =  (3/2) 81  =  243  / 2  =  121.5

Simplify as

2n^2 + 20n - 64 <  0

n^2 + 10n - 32  <  0       set up as an equality

n^2  + 10n  =  32          complete the square on  n

n^2 + 10n + 25  = 32 + 25

( n + 5)^2  =  57      take  both roots

n + 5  =  sqrt (57)      and    n + 5 =  -sqrt (57)

n = sqrt (57)  - 5   ≈ 2.5             n  =  -5 -sqrt (57) ≈ -12.5

These  are roots of the quadratic which opens upward.....every integer between  them will cause the given inequality to be <  0

The number of intgers  =  2  - -12 + 1  =  2 + 12 + 1  =  15

See here : https://www.desmos.com/calculator/8xp9uktmbh

The range ≈   [ -0.22, 0.649  ]

(a + 2) / (a + 1)  =  b /  8

b =   8 (a + 2)  / (a + 1)

The function has a  horizontal asymptote at  b = 8

a          b

0         16

1         12

3         10

7          9

-2        0

-3        4

-5        6

-9        7 

x^2 + 4x - 12  ≤ x^2 + 14

4x  ≤ 26

x  ≤ 26 / 4

x  ≤ 13 / 2

( -inf, 13/2 ]

10^(1/2) / 10^(1/4) =  10 ^(1/2 -1/4)  = 10^(2/4 - 1/4)  = 10^(1/4)

We want all numbers in this form:

_ _ 3

_ _ 4

_ _ 5

An integer can't be part of both groups, since an integer has only one ending number.

For all of these, the first two can cycle from 10 to 99 which is 90 numbers

Sum of numbers ending in 3 is (sum of numbers 10 - 99) * 10 + 90*3 = 49320

Sum of numbers ending in 4 is (sum of numbers 10 - 99) * 10 + 90*4 = 49410
Sum of numbers ending in 5 is (sum of numbers 10 - 99) * 10 + 90*5 = 49500

49320 + 49410 + 49500 = 148230.

Find the sum of the squares of the roots of $2x^2+4x-1=x^2-8x+3$.  

                                           2x² + 4x – 1  =  x² – 8x + 3  

Combine like terms               x² + 12x – 4  =  0  

Complete the square            x² + 12x + 36  =  40  

                                             (x + 6)²  =  40  

                                              x + 6  =  +sqrt(40)  

                                                    x  =  +sqrt(40) – 6  

Square each root      [ +sqrt(40) – 6 ]²  =  40 – 12sqrt(40) + 36      (eq 1)   

                                  [ –sqrt(40) – 6 ]²  =  40 + 12sqrt(40) + 36      (eq 2)   

Add (eq 1) & (eq 2)                                    80                      + 72   

                              Sum of squares  =  152    

_.

To find the value of

x^2+y^2

, we can use the identities:

(x+y)^2=x^2+y^2+2xy

(x3+y3)=(x+y)(x2+y2−xy)

From the given equations, we know that

x+y=2

and

x^3+y^3=5

.

Substituting

x+y=2

into the first identity, we get:

(2)^2=x^2+y^2+2xy

4=x^2+y^2+2xy

We can rearrange this to find an expression for

xy

:

xy=2−(x^2+y^2)/2​

Substituting

x+y=2

and

xy=2−(x^2+y^2)/2

into the second identity, we get:

5=2(2−((x^2+y^2)/2)​−((x^2+y^2​)/20)

5=4−2(x^2+y^2)

Solving for

x^2+y^2

gives us:

x^2+y^2=(4-5)/-2=1/2

So,

x^2+y^2=1/2

edit 1 : im not 100% sure if i'm correct

.

Simplfy as

6x^2 - 21x -31  = 0

Largest  x =   [ 21 + sqrt [ 21^2 - 4*6 * -31 ]  ] / [2 * 6] ≈  4.62

8x^2 + 6x - 1  =  0

ab  = -6/8 = -1/8

a + b  = -6/8 = -3/4          square both sides

a^2 + 2ab + b^2  = 9/16

a^2 + b^2   = 9/16  - 2ab

a^2 + b^2  = 9/16  +1/4

a^2 + b^2  =  13/16

a^3 + b^3  = 

(a + b) ( a^2 + b^2  -  ab)  =

(-3/4) (13/16 + 1/8)  = (-3/4)( 13/16 + 2/16)   (-3/4) ( 15/16)  = -45 / 64

Multiplying both sides by 8(a+1) (which is nonzero since a is an integer), we get: $8(a+2)=b(a+1).$This forces b to be a multiple of 8. We can rewrite the equation as: $b=8k⋅a+1a+2​=8k(1+a+11​)$Since b is an integer, k must also be an integer.

If a=−1, then the fraction becomes undefined, so we must reject this possibility. Otherwise, a+11​ is an integer, so a+1 must divide 1. The only two possibilities are a=0 and a=−2.

If a=0, then b=16k, which means that k can be any integer. So, in this case, there are infinitely many solutions (a,b).

If a=−2, then b=−8k, which means that k must be nonpositive. The only nonpositive integer that divides 8 is −8, so k=−8 and b=64. Thus, the only solution in this case is (−2,64).

In conclusion, there are ∞​ solutions if a=0 and exactly 1​ solution if a=−2.

Absolutely, I’ve been improving my problem-solving abilities in solving polynomial equations. Let's find a3+b3, where a and b are the roots of the equation: 5x2−11x+4=−3x2−17x+5

We can solve the equation for a and b using the quadratic formula and then use the fact that for any quadratic equation ax2+bx+c=0, the sum of the roots is −ab​ and the product of the roots is ac​.

Steps to solve: 1. Solve the equation for a and b: 5x2−11x+4=−3x2−17x+5

Combining like terms and rearranging the equation, we get: 8x2+6x−1=0

Using the quadratic formula, we get: x=2a−b±b2−4ac​​ where a, b, and c are the coefficients of the quadratic equation. In this case, a = 8, b = 6, and c = -1. Substituting these values into the formula, we get:

x=2⋅8−6±62−4⋅8⋅−1​​

x=16−6±217​​

Therefore, the roots are x=817​−3​ and x=8−17​−3​.

2. Find a^3 + b^3: We know that for any quadratic equation ax2+bx+c=0, the sum of the roots is −ab​ and the product of the roots is ac​. In this case, we have:

a+b=−86​=−43​ ab=8−1​

We can use these relationships to find a3+b3:

a3+b3=(a+b)(a2−ab+b2)

Substitute the values we found for a + b and ab:

a3+b3=−43​(a2−ab+b2)

We don't need to find the individual values of a^2 and b^2$ since we can rewrite the expression using the fact (a+b)2=a2+2ab+b2 :

(a+b)2=a2+2ab+b2

Substitute a + b = -3/4:

(−43​)2=a2+2ab+b2

Expand:

169​=a2+2ab+b2

Substitute ab = -1/8:

169​=a2−41​+b2

Combine like terms:

169​+41​=a2+b2

a2+b2=45​

Substitute this back into the expression for a^3 + b^3:

a3+b3=−43​(45​−81​)

a3+b3=−43​⋅89​

a3+b3=−3227​

Answer: a3+b3=−3227​

Absolutely, I’ve been improving my problem-solving abilities in quadratic equations. Let's find the largest value of x such that: 3x2+17x+15=2x2+21x+12−5x2+17x+34

We can solve the equation by combining like terms, moving terms to one side of the equation, and using the quadratic formula.

Steps to solve: 1. Combine like terms: 3x2+17x+15=−3x2+38x+46

2. Move terms to one side: 6x2−21x−31=0

3. Use the quadratic formula: x=2a−b±b2−4ac​​ where a, b, and c are the coefficients of the quadratic equation. In this case, a = 6, b = -21, and c = -31. Substituting these values into the formula, we get:

x=2⋅6+21±(−21)2−4⋅6⋅−31​​

4. Simplify: x=1221±1185​​

Since the discriminant (the part under the radical) is positive, there are two real solutions. However, the question asks for the largest value of x. The larger solution will be the one with the positive square root.

Answer: x=121185​+21​

We can solve the equation for u and v first:

Combine like terms: 7x^2 + 2x - 21 = 0

Factor the equation: (x - 3)(7x + 7) = 0

Therefore, the solutions are x = 3 and x = -1. These correspond to u = 3 and v = -1.

Now, we need to find 1/u^2 + 1/v^2:

Substitute u = 3 and v = -1: 1/3^2 + 1/(-1)^2

Simplify: (1/9) + (1) = 1/9 + 9/9

Combine terms: 10/9

Therefore, 1/u^2 + 1/v^2 = 10/9.

There are two cases to consider for placing the two indistinguishable pieces on the chessboard:

Same Row: The pieces can be placed on any of the 8 rows. Once the row is chosen, either piece can be placed on any of the 8 squares in that row.

So, for the same row scenario, there are 8 ways to pick a row and 8 ways to arrange the pieces within that row, resulting in 8 * 8 = 64 arrangements.

Same Column (excluding diagonal): The pieces can be placed on any of the 8 columns (excluding the possibility of them being on the same diagonal, which we'll address later).

Similar to the same row case, once a column is chosen, either piece can be placed on any of the 8 squares in that column.

However, this counts the diagonal placement twice (once for each way of choosing the row or column first).

So, for the same column scenario (excluding diagonal), there are 7 ways to pick a column and 7 ways to arrange the pieces within that column, resulting in 7 * 7 = 49 arrangements.

Correcting for Diagonal Placement:

We've double-counted the arrangements where the pieces are on the same diagonal.

How many are there? Imagine choosing the diagonal first. There are only 4 choices (one for each corner).

Then, you can place the pieces on two squares out of the 8 on that diagonal, giving 8 choices.

However, since we've already counted this scenario twice (once for each way of choosing the row or column first), we need to subtract 1.

Therefore, the number of diagonal placements counted twice is 4 (diagonals) * (8C2) (combinations to choose 2 squares out of 8) - 1 (double counting) = 14.

Total Arrangements:

Adding the arrangements for same row, same column (excluding diagonal), and subtracting the double-counted diagonals, we get the total number of possible placements:

Total = Same Row + Same Column (excluding diagonal) - Double Counted Diagonals Total = 64 + 49 - 14 Total = 99

There are 99 ways to place two indistinguishable pieces on an 8x8 chessboard if they must be in the same row or the same column.

No bounded region exists

https://www.desmos.com/calculator/uh6sw4gpdm

For how many three-digit positive integers is the sum of the digits equal to 5?  

104      203      302      401      500  

113      212      311      410  

122      221      320  

131      230  

140  

If I didn't miss any, I get 15 of them.    

_.

Laverne starts counting out loud by ${}5$'s.  She starts with $2$.  As Laverne counts, Shirley sums the numbers Laverne says.  When the sum finally exceeds $20,$ Shirley runs screaming from the room.  What number does Laverne say that sends Shirley screaming and running?  

Laverne     Shirley   

       2              2  

       7              9   

     12            21  

_.

No prob.....I have those days when I should have just stayed in  bed, too....