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3x + 2y + xy = 115 + 11x + 2y    simplify

8x - xy = - 115

x ( 8 - y)  =  - 115

x ( y - 8)  =  115

factors of 115  =   1   5   23    115

Four possible solutions  for x, y

y = 31    x =  5

y = 123  x = 1

y = 13    x = 23

y = 9     x = 115

x + y =   36  or  124

a1 + a3 + a5 =  22

a1 + (a1 + d) + (a1 + 2d)   =22

3a1 + 3d =  22   

a2 + a4 =  17

(a1 + d) + (a1 + 3d)  = 17

2a1 + 4d = 17

3a1 + 3d  = 22   mult through by 4  →   12a1 + 12d  =  88

2a1 + 4d =  17   mult through by -3 →     -6a1 - 12d  = -51       add these

6a1  =  37

a1  = 37 / 6

x + y  = 2

square both sides

x^2 + 2xy + y^2  = 4

x^2 + y^2 =  4 - 2xy

xy = 5  -x^2 - y^2

xy = 5 - (x^2 + y^2)

So

x^2 + y^2  = 4 -2xy

x^2 + y^2  = 4 - 2 [ 5 - (x^2 + y^2) ] 

x^2 + y^2  =  4 - 10 + 2(x^2 + y^2)

-3(x^2 + y^2)  =  -6

x^2 + y^2  = -6 / -3  = 2

a + b = 4

square both sides

a^2 + 2ab + b^2  =16  →  a^2 + b^2  = 16 - 2ab

a^2 + b^2 = 6

So

6 = 16 - 2ab

-10 = -2ab

5 = ab

-5 = -ab

a^3 + b^3  =  

(a + b) ( a^2 + b^2 - ab) =

(4) (6 -  - 5)

(4) (11)  = 

44

Let's multiply everything out and see what we have. 

We get 

\(x^3(x^2+tx)\\ x^5+x^4t\)

As shown, there is not x^2 term. 

This means that t can pretty much be anything as long as it doesn't contain x^-2

Thanks! :)

We can convert both numbers to base 10, add them up, and convert them back. 

First, let's convert the two. We have

\((357)_8 = (3 × 8^2) + (5 × 8^1) + (7 × 8^0) = (239)_{10} \)

\((501)_8 = (5 × 8^2) + (0 × 8^1) + (1 × 8^0) = (321)_{10} \)

Adding our two results up, we get the sum

\(321+239=560\)

Now, we must convert this into base 8 again. 

We get \(560_{10}=1060_8\)

So our final answer is 1060. 

Thanks! :)

In order to solve this problem, we can use a very simple trick. 

First, let's set \(x = 0.\overline{2123}\)

If we have this for the value of x, we get \(10000 x = 2123.\overline{2123}\)

Now, we subtract x from 10000x. We get

\(10000x-x= 2123.\overline{2123} -\overline{2123}\\ 9999x=2123\)

This means we get

\(x = \frac{2123}{9999}\)

Simplifying, we get

\(x = \frac{193}{909}\)

So our final answer is \( \frac{193}{909}\)

Thanks! :)

First, let's put s in terms of t so we can do some subsitutions. We get

\(\quad s=-\frac{2\left(-4t+1\right)}{5}\) from the first equation. 

Subsituting this value of s into the second equation, we get

\(3t-6\left(-\frac{2\left(-4t+1\right)}{5}\right)=9-2t\)

Now, we solve for t. We get

\(\frac{-23t+12}{5}=9\)

\(-23t+12=45\)

\(t=-\frac{33}{23}\)

Now, we subsitute t back into the first equation to find s. We get

\(s=-\frac{62}{23}\)

We get \(s=-\frac{62}{23},\:t=-\frac{33}{23}\)

So our final answer is \((-62/23, -33/23)\)

Thanks! :)

We don't have to consider any numbers ending with 5 or even, since we know they are composite numbers. 

,

So now, we go through a list. 

First, start with 2001. We have \(2001=3*667\), and it has 8 factors, so it is composite. 

2003 is prime, as it is only divisible by 1 and itself. 

We have \(2007=3*669\), so 2007 is composite. 

\(2009 = 7*287\), so 2009 is composite. 

\(2011\) is prime. 

\(2013=671*3\), so 2013 is composite.

\(2017 \text{and} 2027\) are prime, but everything in between is composite. 

\(2029\) is prime, so that's our final answer. 

So 2029 is our answer. 

Thanks! :)

A prime number is a number only divisble by 1 and itself. 

Let's take a look at the numbers we have. 

First, we have \(1\). It is only divisble by 1 and itself, but it only has one true factor since \(1=1*1\), so it is not consdered prime. 

Next, we have 2. It is divisble as \(2 = 2*1\), so it is prime. 

Going through the list, \(3= (3*1), 5 = (5*1), 7=(1*7)\) are all prime numbers. 

However, \(4 = (2*2), 6 = (2*3), 8 = (2*4), 9=(3*3), 10=(2*5)\)are all composite numbers. 

So our final answer is \(2,3,5,7\)

Thanks! :)

First, let's subtract the to numbers and see what we get. 

We get \(123 - 47 = 76\)

Now, we need to find the prime factors of 76. 

When we factorize it, we get \(76 = 2^2 \cdot 19\)

This means there are only two distinct prime factors. \(2, 19\)

Adding these two up, we get \(19+2=21\)

So 21 is our final answer!

Thanks! :)

Yall were both so helpful thank you.

Simplify as

8a + b = -3

9a + (2 +m)b = k 

This will have infinte solutions when the equations are the  same...so....multiply the first equation  by 9 and the  second by 8 and  we  get

72a + 9b = -27

72a + 8 (2 + m)b = 8k

So

9 = 8(2 + m)            and          8 k  = -27

9 = 16 + 8m                              k = -27/8                           

-7 = 8m

m = -7/8

1.  ax^2 + bx + a

Product of the roots =  a/a   = 1

So....call the second root, m

And

m * 4  = 1

m =  1/4 = the other root

Let 2.57....  = x

Let  257.57... = 100x

100x   =   257.57....

-    x       -   2.57.....   

__________________________

99x  =     255

x = 255 / 99 =  85 / 33

The answer is E075A.

The answer is 16/93.

N = 407.

Problem 2

Note that both factors share some common terms. Let's rewrite them to group these terms together:

(x^50 + x^48 + ... + x^2 + 1) + (x^49 - x^47 + ... + x - 1)

According to the sum of a finite geometric series formula, the sum of each individual sequence is:

(x^50 + x^48 + ... + x^2 + 1) = (x^(50)) * (1 / (1 - x^2)) = x^(50) / (1 - x^2)

(x^49 - x^47 + ... + x - 1) = (x^(49)) * (1 / (1 + x^2)) = x^(49) / (1 + x^2)

Now we can find the product:

Product = (x^(50) / (1 - x^2)) * (x^(49) / (1 + x^2))

Since we're only interested in the sum of the coefficients, we can ignore the constant terms in the denominators (they won't affect the sum). This simplifies the product to:

Product ≈ x^(50 + 49) // Ignoring constant terms in denominators

The sum of the coefficients in this product is simply the coefficient of the term with x raised to the highest power, which is x^(99). The coefficient of this term is 1.

In conclusion, the sum of the coefficients of the product is 1.

The answer is 17/210.

The answer is 40568/7.

The answer is 51485016_16.

Problem 1:

We can leverage the concept of sum and product of roots in quadratic equations.

Since x = 4 is a root, we know that f(4) = 0. This translates to:

16a + 4b + a = 0

Combining like terms, we get: 17a + 4b = 0 (Equation 1)

We are also given that f(x) = ax^2 + bx + a. Since a ≠ 0, we can rewrite f(x) by factoring out a:

f(x) = a(x^2 + bx/a + 1/a)

We know that one root is x = 4. This means that (x - 4) is a factor of f(x). Let's rewrite f(x) using this information:

f(x) = a(x - 4)(x + r) for some constant r (since the product of the roots is cr)

Equating both ways of expressing f(x), we get:

a(x^2 + bx/a + 1/a) = a(x - 4)(x + r)

Expanding both sides:

ax^2 + bx + a = ax^2 - (4a + r)x + 4ar

Equating the coefficients of like terms on both sides, we get:

b = - (4a + r) (Equation 2)

Now we have two independent equations (Equation 1 and Equation 2) with unknowns a, b, and r. We can use the fact that one root is x = 4 (reflected in Equation 1) to solve for the other root.

Since we're looking for the other root, let's try to eliminate a and b from the equations and solve for r.

From Equation 1: b = -17a/4

Substitute this into Equation 2:

-17a/4 = - (4a + r)

Solve for r:

17a/4 = 4a + r

r = 17a/4 - 4a

r = a/4

Now that we know r (product of roots), we can find the other root itself. Since the product of roots (cr) is also equal to f(4) (which is 0 in this case), we have:

(4)(other root) = 0

Since a ≠ 0, this implies the other root must be:

Other root = 0

In conclusion, the other root of the equation f(x) = 0 is x = 0.

Analyzing f(x) and Asymptotes:

1. Vertical Asymptotes:

f(x) = 0 when x = 0. Since dividing by zero is undefined, this means there's a vertical asymptote at:

x = 0

2. Horizontal Asymptotes:

The degree of the numerator (1) is less than the degree of the denominator (2) in f(x). Following the first case mentioned earlier:

The horizontal asymptote of y = 1/(f(x)) is y = 0 (the x-axis).

3. Oblique Asymptotes:

The degree of the numerator isn't one greater than the degree of the denominator. Therefore:

There are no oblique asymptotes.

In conclusion:

Vertical asymptote: x = 0

Horizontal asymptote: y = 0

A = 10    B = 11

E = 14

2 (12)^3  + 10  (12)^2  + 3( 12 ) + 1  = 4933

14 * 12  + 2 =  169

4933 + 169  = 5102

5102 / 12  = 425 R 2

425 / 12 = 35 R 5

35 /12  = 2 R 11

2/12 = 0 R 2

2B52₁₂