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We start by simplifying

\(4x^2+14x-c-3x^2-12x\)

\(x^2+2x-c\)

To find all values of c, we take 2x the middle term, divide by 2, then x, which gives us 1, 

So, a square root of some number should equal 1, which and that number is one, so -c = 1, c = - 1 

This may sound complicated, dividing by so many numbers, but it's really not, the reason we do that

is because when (ax+b)² is expanded, you get ax²+2abx+b² so to get b when provided the first term and the middle term, we divide by 2ax, and square the result. 

We have:

\(\frac{x}{x-5}=\frac{4}{x-4}+\frac{12}{x^2-9x+20}\)

The best way to approach this is to get rid of the fractions.

We notice that x²-9x+20 = (x-4)(x-5)

Now we multiply both sides by x²-9x+20

\((x^2-9x+20)(\frac{x}{x-5})=(x^2-9x+20)(\frac{4}{x-4}+\frac{12}{x^2-9x+20})\)

This equals:

\(x(x-4)=4(x-5)+12\)

\(x^2-4x=4x-20+12\)

\(x^2-8x+8=0\)

We notice this is not factorable, so we use the quadratic formula:  (+- means plus minus)

\({x}=\frac{8+-\sqrt{32}}{2}\)

\({x}_{1,2}=4+2\sqrt{2}, 4-2\sqrt{2}\)

1 - [(3/12) (2/11) (1/10)] = 1 / 220

2 - [+ 3 (3/12) (2/11) (9/10)] = 27 / 220

3 - [+ (1/8 - (1/8) (3/12) (2/11) (1/10))] = 219 / 1760

4 - [- (1/8) 3 (3/12) (2/11) (9/10)] = - 27 / 1760

5 - The probability is: [1 / 220  +  27 / 220  +  219 / 1760]  -  27 / 1760 = 13 / 55

11 / 10 * 12 / 11 * 13 / 12 * 14 / 13 * 15 /14 * 16 / 15 * 17 / 16 * 18 / 17 * 19 / 18 * 20 / 19 * 21 / 20 * 22 / 21* 23 / 22 * 24 / 23 * 25 / 24 * 26 / 25 * 27 / 26 * 28 / 27 * 29 / 28 * 30 / 29 ==3 

The first fraction is:  11 / 10

The first fraction is 2/7.

The answer is (-inf, -2].

The only two digit numbers divisible by 19 or 31 are, 19, 38, 57, 76, 95 (those by 19), and 31, 62, 93, (by 31).

If the first digit is 1, the second digit has to be 9, (19  being the only possible continuation).

The sequence could then continue with either 95 or 93, forming 195 or 193 respectively.

Continuing in this way, the only possibles are,

195762, (19), (95), (57), (76), (62), the sequence terminating.

1938, (19), (93), (38), the sequence terminating.

1931931931.....(19), (93), (31), (19), (93), .....

This third sequence can continue  indefinitely or it could be arranged to terminate by attaching either a 5 after a 9 or an 8 after a 3.

To reach 2002  digits, we have to go 193193193193...and, if we continue in this way, the 3 would land on the 2001th digit meaning the the 2002nd digit would be a 1.

However the sequence could be broken one cycle earlier, terminating with 1938 or 1957(62),

So 8 is the largest possible digit.

Use power of a point, and please copy and paste the diagram over, since we do not have a visual.

The answer is( -2.828,0) or (2.828,0). Either way, they are the same distance from the origin.

I answered this same question a few days ago, here's the link:

f(n) = 1/(n^2 + 4n + 5)

c = 488.

By the maximum modulus principle, the answer is 4.

Part B)
We can prove this by using the trivial inequality:

\((a-b)^2>=0\)

We expand

\(a^2-2ab+b^2>=0\)

Rearrange

\(a^2+b^2>=2ab\)

Doing this with all the variables gets us \(c^2+d^2>=2cd\) and \(e^2+f^2>=2ef\)

Multiplying these inequalities gets us, \((a^2+b^2)(c^2+d^2)(e^2+f^2)>=8abcdef\)

Note: We can multiply these inequalities because they are positive, and they have the same signs.

The answer is 28.

The constant term is -18.

114th term of:   (2u - 3v + u^2 - v^2)^9==- 577368 u^2 v^9

There are 4 right triangles as follows:

1 = (5, 13)
2 = (9, 15)
3 = (16, 20)
4 = (35, 37)

The 3rd side being 12 in each of above 4 triangles.

(4x - 7y)^n. 

n cannot be 2, because you will get an xy term, and x^2 term, and a y^2 term with coefficients.

From binomial theorem logic, you can see that cx^a*y^b, where a = b + 1, and a + b = n, a = 2, so we have a + b = 3, and n = 3.

Nice! Thanks for double-checking TooEasy's question! 81/4 is the answer :D

We can list out the first few digits.

1 is the first, followed by a 9 to make the first constraint true.

Then, the 9 has to have a following 3 because 93/31 = 3. (So far is 193____)

Then after the 3 there is an 8 or 5 to make the constraint true. (So far is 1938______ or 1935______)

If 8 then impossible because no 2 digit number that starts with 8 is divisible by 31 or 19. So 1935 is accurate, after the 5 there is a 7 because 57 is divisible by 19. (So far 19357______)

After a 7 there is a 6, because 76/19 = 4. No 2 digit number with 7s as tens digit is divisible by 31 so 31 is out of this digit part. (So far is 193576___).

After the 6 there is a 2, 62/31 = 2. No 2 digit number with 6 as tens digit is div by 19. (So far is 1935762____).

After the 2, there is nothing, so unfortunately we have to start over to step 3, where instead after the 9 there is a 5. (So far is 195_____)

After the 5 is a 7. (So far is 1957_____) After the 7 there is a 6 (1957) After 7 is a 6, then after 6 is a 2. So it loops back. So whoever made this problem forgot to realize there is no such string, since no 2 digit number that starts with a 2 or 8 is divisible by either 31 or 19 :|

zw - 3w - 2iw + 4z = 7iw + 5w + 11iz - 13 + 15i

z = a + bi, w = c + di

(a + bi)(c + di) - 3(c + di) - 2i(c + di) + 4(a + bi) = 7i(c + di) + 5(c + di) + 11i(a + bi) - 13 + 15i

ac - bd + (bc + ad)i - 3c - 3di - 2ci + 2d + 4a + 4bi = 7ci - 7d + 5c + 5di + 11ai - 11b - 13 + 15i

Combine like terms.

[ac - bd - 3c + 2d + 4a] + [bci + adi - 3di - 2ci + 4bi] = [-7d + 5c - 11b - 13] + [7ci + 5di + 11ai + 15i]

Set Re = Re2, and Im = Im2

ac - bd - 3c + 2d + 4a = -7d + 5c - 11b - 13

ac - bd + 4a + 11b - 8c + 9d = -13

bci + adi - 3di - 2ci + 4bi = 7ci + 5di + 11ai + 15i

bc + ad - 3d - 2c + 4b = 7c + 5d + 11a + 15

ad + bc - 11a + 4b - 9c - 8d = 15

Notice how many of the coefficients seem swapped. This might imply something about conjugates. GL (you do it now)

I think the problem had a typo. 

x^2 - 4x = 4x + 20

it gives imaginary, perhaps the owner of the question meant the right hand side to be 5/(x - 4)