All Answers 356323 Answers

The quadratic equation takes the form of x^2 - mx + 24 = 0. Given that x_1 and x_2 are its roots and are integers, it means that the sum of these roots (x_1 + x_2) equals m (based on Vieta's formulas), and the product of these roots equals 24 (x_1*x_2=24). In this case, we need to factor the constant term (24), in other words, list all possible pairs of integers that multiply to give 24. These include {1,24}, {-1,-24}, {2,12}, {-2,-12}, {3,8}, {-3,-8}, {4,6}, {-4,-6}, and their negative counterparts. As both roots are integers, and each pair of factors correspond to one unique value of m (the sum of those two integers), there are altogether 16 different values for m, considering both positive and negative pairs.

Answer is: 16

That is not the correct answer, can you check your explanation again please? Thank you

https://web2.0calc.com/questions/quadratic_30479 

                           A

                 10               15

         B      1        D      3         C 

Let DC =  3BD

AD =  sqrt [ AC^2  - (3BD)^2 ]  = sqrt [ 15^2 - 9BD^2 ]      (1)

AD = sqrt [ AB^2 - BD^2 ]  =  sqrt [ 10^2 - BD^2 ]       (2)

Equate (1) , (2)

sqrt [ 15^2 - 9BD^2 ] = sqrt [ 10^2 -BD^2 ]              square both sides

15^2 - 9BD^2 = 10^2 - BD^2

225 - 9BD^2 = 100 - BD^2

125 = 8BD^2

125/8  = BD^2

AD = sqrt [ 10^2 - BD^2]  = sqrt [ 100 - 125/8 ] = sqrt [ 675/8]  = [15sqrt 3 ]  / [2 sqrt 2] =

7.5 sqrt [3/2]  =   7.5 sqrt (6) / 2   =  3.75sqrt (6) ≈ 9.19

                             A

                                     8

             B     2       K      3       C

Since AK is an  altitude

AK = sqrt [ AC^2 - KC^2 ]   =  sqrt [ 8^2  - 3^2 ] = sqrt [ 55]

And

AB = sqrt [ AK^2  + BK^2 ]  =  sqrt [ (sqrt 55)^2 + 2^2 ] = sqrt [ 59]

Sorry, there is a mistake in my argument: \(\frac n2\) can in fact be 729, so the lower bound is \(n \geq 1458\). The answer is then 365 instead of 364.

Note that since the rectangles are congruent, we have DB = BF. That would imply \(\angle BDF = \angle BFD\). Also, BE is perpendicular to EF, so \(\angle BED = \angle BEF = 90^\circ\). Together with the common side BE we have the pair of congruent triangles \(\triangle DEB \cong \triangle FEB\). Consequently we have the fact that E is the centre of the semi-circle. So we now have enough information to figure out that \(\triangle DAE\) is equilateral. The rest is done through some angle chasing, which is not hard.

You are given the dimensions of the rectangle directly, the area is of course \(WX \cdot XY = 2 \cdot 3 = 6\).

Let \(CX = k\). Then, since AX is parallel to BY, we have

\(\tan^{-1} \dfrac{k}{15} + \tan^{-1} \dfrac{3k}{20} = 60^\circ\)

Taking tan on both sides and using compound angle formula,

\(\dfrac{\frac k{15} + \frac{3k}{20}}{1 - \frac{k}{15} \cdot \frac{3k}{20}} = \sqrt 3\\ \dfrac{\sqrt 3 k}{45} + \dfrac{\sqrt 3 k}{20} = 1- \dfrac{k^2}{100}\\ \dfrac{13\sqrt 3k}{180} = 1 - \dfrac{k^2}{100}\\ k^2 - 100 = -\dfrac{65 \sqrt 3 k}9\\ k^2 + \dfrac{65 \sqrt 3}9k - 100 = 0\\ \left(k + \dfrac{65 \sqrt 3}{18}\right)^2 = 100 + \left(\dfrac{65 \sqrt 3}{18}\right)^2 = \dfrac{15025}{108}\\ k = -\dfrac{65 \sqrt 3}{18} \pm \dfrac{5 \sqrt {601}}{6 \sqrt 3} \text{(reject negative)}\\ k = -\dfrac{65 \sqrt 3}{18} + \dfrac{5 \sqrt{1803}}{18}\)

Hence, by Pythagoras' theorem, 

\(AB = \sqrt{(20 - 15)^2 + k^2} = \dfrac{5}{18} \sqrt{2634 - 78 \sqrt{601}}\)

I think a slight problem to your solution is that you are saying our (X,Y) can be equal to 5 but the quesiton doesn't want our pieces to even equal to five. Would this change your solution? I haven't entered anything. I wanted to check before doing any submitting. 

Thanks for the help!

A natural question to ask is: Does the inverse exist?

This is the graph of the y = f(x). Obviously, the graph does not pass the horizontal line test, and hence is not 1-1. 

The inverse function does not exist, unless you restrict the domain even more.

Suppose X and Y are random variables which follow the continuous uniform distribution on [0, 6]. These random variables will represent the points where we break the stick at.

The probability in question is simply expressible as \(\mathcal P (\min(X, Y) \leq 5\text{ and }\max(X, Y) \geq 1\text{ and }|X - Y| \leq 5)\).

Let \(S = \{(x, y): 0\leq x \leq 6, 0 \leq y \leq 6\}\) and \(P = \{ (x, y) \in S\mid\min(x, y) \leq 5\text{ and }\max(X, Y) \geq 1\text{ and }|X - Y| \leq 5\}\). The required probability is \(\dfrac{\text{Area}(P)}{\text{Area}(S)}\). The following is a graph of P:

Now, the probability is \(\dfrac{\text{Area}(P)}{\text{Area}(S)} = \dfrac{(5 - 1)^2}{(6 - 0)^2} = \dfrac49\).

Using Law of Cosine in \(\triangle PQR\) gives

\(8^2 = 5^2 + 11^2 - 2(5)(11) \cos \angle Q\\ \cos \angle Q = \dfrac{5^2 + 11^2 - 8^2}{2(5)(11)}\\ \cos \angle Q = \dfrac{41}{55}\)

Using Law of Cosine in \(\triangle PQM\) gives

\(PM^2 = 5^2 + \left(\dfrac{11}2\right)^2 - 2(5)\left(\dfrac{11}2\right) \cos \angle Q\\ PM^2 = 5^2 + \left(\dfrac{11}2\right)^2 - 2(5)\left(\dfrac{11}2\right)\left(\dfrac{41}{55}\right)\\ PM^2 = \dfrac{57}4\\ PM = \dfrac{\sqrt{57}}2\)

Suppose \(\overline{AC} = x\) and \(\overline{BC} = y\). From \(\overline{AM} = \overline{BN} = 1\), we have

\(\begin{cases} \left(\dfrac x2\right)^2 + y^2 = 1\\ x^2 + \left(\dfrac y2\right)^2 = 1 \end{cases}\)

Adding, we have 

\(\dfrac 54 \left(x^2 + y^2\right) = 2\\ x^2 + y^2 = \dfrac 85\\ \sqrt{x^2 + y^2} = \dfrac{2 \sqrt {10}}5\)

Hence, length of the hypotenuse is \(\dfrac{2 \sqrt {10}}5\)

The most important thing is the unit of n. Simply compute by law of indices:

\(n = \dfrac ky = \dfrac{1\text{ kg m}^{1}}{1\text{ kg m}^{-1}} = 1\,\text{m}^2\)

Thanks very much

1. Do this: \(\text{There are 5 available options: xxxxx.} \\\text{These are the possible combinations: 24xxx, x24xx, xx24x, and xxx24} \\\text{it follows that: } \\10\cdot10\cdot10=1000\\ 9\cdot10\cdot10-1=899\\ 9\cdot10\cdot10-1=899\\ 9\cdot10\cdot10-2=898\\ \text{The answer is } 3696\)

The other condition seems to be that you specify in advance the values of x, y and z.  It wasn't clear from your original question that that was what was intended!

a problem would be something like a+b+c=4,ab+ac+bc=22, abc=65. I don't really understand what you ment by an infinite number of solution, but thanks for taking the time to answer.

3.   Angle  B  = 180 - 55 -26  = 99°

10 / sin 99 = BC / sin  55

BC =  10sin55 / sin 99    ≈  8.3

2.

Call the point where the diagonals meet =  O

Triangle SOR  similar to triangle XOY

XY  / SR  = 5/28

So...height of triangle XOY  to height of triangle SOR = 5/28

Then there are 5 + 28  =  33  equal parts  in 1/2 height of the trapezoid

And triangle POQ similar to  triangle SOR

And height of POQ to height of SOR =    [ 5 +33] / 28  =  38/28   = 19/14

So

PQ  =  (19/14)SR=  (19/14) * 28 =   38 

1.

tan x + sec x =  2

sinx / cos x  + 1/cos x  = 2

sinx + 1   = 2cos x           square bith sides

sin^2 x + 2sin x + 1  = 4cos^2 x

sin^2 x + 2sin x + 1 =  4 (1-sin^2 x)

sin^2 x + 2sin x + 1 =  4 - 4sin^2 x

5sin^2 x + 2sin x - 3 =   0

(5sin x -3  )(sinx + 1)   = 0 

5sin x - 3  =  0                      sin x + 1  =  0

5sinx  = 3                              sin x  = -1

sin x =  3/5                                x = -pi/2     reject

cos x =   sqrt [ 5^2 - 3^2 ] / 5  =  sqrt [ 16 ] / 5  =   4 / 5

\(\begin{cases} x^2 + y^2 = 5\\ y= \dfrac{3x+2}4 \end{cases}\)

Substituting, we have

\(x^2 + \left(\dfrac{3x + 2}4\right)^2 = 5\\ \dfrac{25}{16}x^2 + \dfrac34x - \dfrac{19}4 = 0\\ 25x^2 + 12x - 76 = 0\\ (x + 2)(25x - 38) = 0\\ x = -2 \text{ or }x = \dfrac{38}{25}\)

When x = -2, y = (3(-2) + 2)/4 = -1.

When x = 38/25, y = (3(38/25) + 2)/4 = 41/25.

The required distance is:

\(\sqrt{\left(-2 - \dfrac{38}{25}\right)^2 +\left(-1 - \dfrac{41}{25}\right)^2 } = \dfrac{22}5\) units.