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The polynomial f(x) has degree 3. If f(-1)=15, f(0)=0, f(1)=-5, and f(2)=12, then what are x-intercepts of the graph f?

\(f(x)=ax^3+bx^2+cx+d\)

\(f(0)=0\qquad so \qquad d=0\\ f(x)=ax^3+bx^2+cx\)

\(also\\ f(-1)=15,\;\;  f(1)=-5, \;\;and \;\; f(2)=12\)

so

\(f(-1)=-a+b-c=15\qquad(1)\\ f(1)=a+b+c = -5\qquad\quad(2)\\ f(2)=8a+4b+2c=12\qquad(3)\\ (1)+(2) 2b=10\\ b=5\\\)

Now we have

\(-a+5-c=15\qquad(1)\\ a+c=-10 \qquad \qquad (1b)\\ \\~\\ 8a+20+2c=12\qquad(3)\\ 4a+10+c=6 \\ 4a+c=-4 \qquad \qquad(3b)\\ \\~\\ (3b)-(1b)\\ 3a=6\\ a=2\\ c=-12 \)

\(f(x)=ax^3+bx^2+cx\\ f(x)=2x^3+5x^2-12x\\ f(x)=x(2x^2+5x-12)\\ f(x)=x(2x^2+8x-3x-12)\\ f(x)=x[2x(x+4)-3(x+4)]\\ f(x)=x[(2x-3)(x+4)]\\ \text{Roots are x = 0, 1.5 and -4}\)

Sounds like a very good guess to me. :)

\(f(x) = \dfrac{a x + b}{x + c} \\ \\ \text{there is a vertical asymptote at }x=3 \text{ so clearly c=-3} \\ \\ \lim \limits_{x \to \pm \infty} f(x) = a \text{ so }a=-4 \\ \\ f(x) = \dfrac{-4x + b}{x-3} \\ \\ f(1) = \dfrac{-4+b}{1-3} = 0 \\ \\ \text{so clearly b=4} \text{ and thus }f(x) = \dfrac{-4x+4}{x-3}\)

see if you can work out part (b) using this as a template

A real number x is chosen at random such that -2 < x < 5. What is the probability that x^2 < 2? 

if x^2 <2

then

\(-\sqrt2

The distance between these is  \(2\sqrt2\)

The distance between -2 and 5 is 7

so

The prob will be   \(\frac{2\sqrt2}{7}\)

it won't be true as long as \(P(A) \neq P(B)\)

\(P(A \wedge B) = P(A|B)P(B) = P(B|A)P(A) \\ \\ \text{clearly } P(A) \neq P(B) \Rightarrow P(A|B) \neq P(B|A)\)

\(\begin{align*} &-2c-2a-3y-2b-3x = \\ \\ &-2(a+b+c) - 3(x+y) = \\ \\ &-2(-3) -3(-9) = \\ \\ &6 -(-27) = 33 \end{align*}\)

If a + b = 9 and x + y = 1,

what is -3b - 3y - 3a - 3x?

\(\begin{array}{|lrcll|} \hline & a+b &=& 9 \\ & x+y &=& 1 \\ \text{sum} & a+b+x+y &=& 9+1 \\ & a+b+x+y &=& 10 \\\\ & && -3b - 3y - 3a - 3x \\ & &=& -3\cdot (b+y+a+x) \\ & &=& -3\cdot (a+b+x+y) \quad & | \quad a+b+x+y = 10 \\ & &=& -3\cdot (10) \\ & &=& -30 \\ \hline \end{array}\)

Hi guest,

yep, I got the answer as soon as I got home yesterday...thanx for your time!!

Please help!

Solve the system of equations by elimination:

\(\begin{array}{|lrcll|} \hline (1) & 4x + 8y &=& -16 \quad & | \quad -8y \\ & 4x &=& -16 -8y \\\\ (2) & -8x + 10y &=& -20 \\ & -2\cdot (4x) + 10y &=& -20 \quad & | \quad 4x = -16 -8y \\ & -2\cdot ( -16 -8y) + 10y &=& -20 \\ & 32 + 16y + 10y &=& -20 \\ & 32 + 26y &=& -20 \quad & | \quad -32 \\ & 26y &=& -20-32 \\ & 26y &=& -52 \quad & | \quad : 26 \\ & y &=& -\frac{52}{26} \\ & \mathbf{y} & \mathbf{=} & \mathbf{-2} \\\\ & 4x &=& -16 -8y \quad & | \quad :4 \\ & x &=& -4 -2y \quad & | \quad \mathbf{y = -2} \\ & x &=& -4 -2\cdot (-2) \\ & x &=& -4 +4 \\ & \mathbf{x} & \mathbf{=} & \mathbf{0} \\ \hline \end{array}\)

Somebody please help!

Here is my best guess! I think it is 964 and the common ratio is 2/3.

9 x 2/3 =6 x 2/3 = 4

Why are you having difficulty with this?

First thing you should do is to expand the terms of each sequence until you exceed 300!!

Sequence A: 2, 4, 8, 16, 32, 64, 128, 256, 512.......etc.

Sequence B: 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, 220, 240, 260, 280, 300, 320.....etc.

Then, start with the first term of sequence A, which is 2. Can you subtract it from ANY term of sequence B and get a POSITIVE DIFFERENCE? Obviously not!. How about the 2nd term, which is 4? The same thing applies to the second term. Try the 3rd term which is 8. Same thing! 16 same thing! Now, we get to 32. Obviously you can subtract 32 from the smallest term in sequence B, which happens to be the first term which is 20. So, 32 - 20 =12. But is it the LEAST POSITIVE DIFFERENCE? Well, you can find out by checking the rest of the terms. The next is 64. Now you can subtract it as follows: 64 - 20 =44, 64 -40 =24, 64 -60 =4, which appears to be the LEAST POSITIVE DIFFERENCE!!!

If you try the next term which is 128 and subtract it from the first 6 terms of sequence B, you will get a positive difference greater than 4. The same applies to 256. And that is all.

Someone please help!!!

Never mind, I found it

I'm sorry, but I am still confused. What about other numbers in the range, like say 947?

Recurrence relation:
a(n + 1) = 2 a(n) (for all n>=1)
1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192.........etc.
Now, you can see the answer for yourself.

I got V=16 and B=50 just gonna go with that

I understand, somewhat. I knew how to do that part but not what comes after..

8*146 and 8*6

12V+1168-48V=592

(I'm using V and B i hope that's okay!)

Then what do I do next with the 12V and 48V?

Well, suppose we want to check and see if 923 is a prime number or not. You take the square root of 923 and you get about 30.38. So, if you want to check 923 to see if it is prime or not, all you have to do is to check only prime divisors under 30.38, which means dividing 923 by all prime numbers from 2 to 30.38. So, 923/2, 923/3, 923/5........all the way to 923/29.

29 is the last prime below 30.38 to check. If 923 doesn't divide any of these primes EVENLY, then it is a prime number. As it happens, this number of 923 is NOT prime because it is divisible by 13.

So, 923/13 =71.

No solution?

It's a parallelogram, so \(2x+2y=180\)

Dividing by 2, we get \(x+y=90\), and so \(AED=180-90=90\)

10 is the circles DIAMETER, not radius.

So it's just \(10*\pi\)

As they are independent

P[rain on Saturday AND rain on Sunday] = P[rain on Saturday] P[rain on Sunday] = (0.6)(0.25) = 0.15 = 15%