anonymous
  • anonymous
Medal to helper! Can you please check my answers? Important! 5Q's. Thanks!
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1. What is the slope of the line which passes through (−2, 0) and (0, 4)? A. −2 B. 0 C. 2 <--- My Answer D. Undefined
anonymous
  • anonymous
2. Use the table below to answer this question: x y −1 7 3 3 5 1 Find the average rate of change for the given function from x = −1 to x = 5. A. −6 B. −1 <--- My Answer C. 1 D. 6
anonymous
  • anonymous
3. What is the equation in point−slope form of the line passing through (0, 6) and (1, 3)? A. (y − 3) = −3(x − 1) <--- My Answer B. (y + 3) = 3(x + 1) C. (y + 3) = −3(x + 1) D. (y − 3) = 3(x + 6)

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anonymous
  • anonymous
4. What is the equation of the graph below? A. y = − (x + 2)^2 + 2 B. y = − (x − 3)^2 + 2 C. y = (x − 2)^2 + 2 D. y = (x + 3)^2 + 2 <--- My Answer
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anonymous
  • anonymous
5. Simplify (x − 4)(x^2 + 3x + 2). A. x^3 − x^2 + 14x − 8 B. x^3 + 7x^2 − 10x − 8 C. x^3 − x^2 − 10x − 8 D. x^3 + 7x^2 + 14x − 8 <--- My Answer
anonymous
  • anonymous
@phi @Hero @SolomonZelman @tom982
SolomonZelman
  • SolomonZelman
#1 Correct #2 Correct #3 Wrong #4 Wrong #5 Wrong
SolomonZelman
  • SolomonZelman
No, #3 might be right, let me check again
SolomonZelman
  • SolomonZelman
(0, 6) and (1, 3) slope=m=(6-3)/(0-1)=3/(-1)=-3 So, with slope -3 and point (1,3) you have: y - 3 = -3•(x-1) So #3 is correct. Alternatively, you could have used the other point (0,6) y - 6 = -3•(x-0) y - 6 = -3x
SolomonZelman
  • SolomonZelman
yes, #3 is correct, but #4 and #5 are for sure wrong.
anonymous
  • anonymous
Okay thanks. Could you help me with those two?
SolomonZelman
  • SolomonZelman
Yes, surely. the general form of the equation of the parabola with vertex at (h,k) is: \(\large\color{#000000 }{ \displaystyle y=a(x-h)^2+k }\) If the parabola opens up, a>0 if the parabola opens down, a<0
SolomonZelman
  • SolomonZelman
So, your coefficient must be negative, because the parabola opens down.
SolomonZelman
  • SolomonZelman
And, tell me where is the center of your parabola located?
SolomonZelman
  • SolomonZelman
I mean the vertex, not the center
anonymous
  • anonymous
(3,2)
SolomonZelman
  • SolomonZelman
Yes, that is right.
anonymous
  • anonymous
So, y = − (x − 3)^2 + 2 is correct, right?
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle y=a(x-h)^2+k }\) h=3 k=2 \(\large\color{#000000 }{ \displaystyle y=a(x-3)^2+2 }\) So the closes option is \(\large\color{#000000 }{ \displaystyle y=-(x-3)^2+2 }\)
anonymous
  • anonymous
Alright thank you. Now number 5.
SolomonZelman
  • SolomonZelman
If you want, also, you can solve for "a", and get the answer without eliminating the rest of the options. Want to solve for a, or to proceed to #5 right away?
anonymous
  • anonymous
I would like to move on to #5.
SolomonZelman
  • SolomonZelman
Alright ...
SolomonZelman
  • SolomonZelman
You have to expand: \(\large\color{#000000 }{ \displaystyle (x-4)(x^2+3x+2) }\)
SolomonZelman
  • SolomonZelman
Can you tell me what do you get if you were to expand \(\large\color{#000000 }{ \displaystyle (x-4)\cdot x^2 }\)
SolomonZelman
  • SolomonZelman
You can use the following rules \(\large\color{#000000 }{ \displaystyle (a+b)\times c=(a\times c)+ (b\times c) }\) \(\large\color{#000000 }{ \displaystyle (a-b)\times c=(a\times c)- (b\times c) }\)
anonymous
  • anonymous
multiply x^2 and x, then multiply x^2 and -4?
SolomonZelman
  • SolomonZelman
yes, and wha do you get after doing this?
anonymous
  • anonymous
x^3 - 4x^2
SolomonZelman
  • SolomonZelman
yes, very good
SolomonZelman
  • SolomonZelman
Now, what do you get after expanding \(\large\color{#000000 }{ \displaystyle (x-4)\times 3x }\)
anonymous
  • anonymous
3x^2 - 12x
SolomonZelman
  • SolomonZelman
Yes, very good
SolomonZelman
  • SolomonZelman
And lastly what do you get after expanding \(\large\color{#000000 }{ \displaystyle (x-4)\times 2 }\)
anonymous
  • anonymous
2x - 8
SolomonZelman
  • SolomonZelman
Yup, now we will put the peaces together...
SolomonZelman
  • SolomonZelman
We wanted to expand \(\large\color{#000000 }{ \displaystyle (x-4)(x^2+3x+2) }\) And this is the same as expanding \(\large\color{#000000 }{ \displaystyle (x-4)\cdot x^2 }\) \(\large\color{#000000 }{ \displaystyle (x-4)\cdot3x }\) \(\large\color{#000000 }{ \displaystyle (x-4)\cdot2 }\) separately, and then adding the results. We get that: \(\large\color{#000000 }{ \displaystyle (x-4)\cdot x^2=\quad +x^3-4x^2 }\) \(\large\color{#000000 }{ \displaystyle (x-4)\cdot3x=\quad +3x^2-12x }\) \(\large\color{#000000 }{ \displaystyle (x-4)\cdot2 =~{{\tiny~~}}\quad +2x-8}\) And add the results: \(\large\color{#000000 }{ \displaystyle x^3-4x^2+3x^2-12x+2x-8 }\)
SolomonZelman
  • SolomonZelman
Can you simplify \(\large\color{#000000 }{ \displaystyle x^3-4x^2+3x^2-12x+2x-8 }\) ?
anonymous
  • anonymous
Yes, let me try.
SolomonZelman
  • SolomonZelman
go ahead and take your time ;)
anonymous
  • anonymous
x^3 - x^2 - 10x - 8
SolomonZelman
  • SolomonZelman
yup, and that is your final answer.
anonymous
  • anonymous
Wow. I would have never guessed that one. You're explanation makes sense. Thank you so much!
anonymous
  • anonymous
Could you check this too?: Which of the following is the conjugate of a complex number with 2 as the real part and −8 as the imaginary part? A. −2 + 8i B. 2 + 8i <--- My Answer C. 2 − 8i D. −2 − 8i
SolomonZelman
  • SolomonZelman
|dw:1449680605870:dw|
anonymous
  • anonymous
Gotcha.
SolomonZelman
  • SolomonZelman
You number is: 2-8i So the conjugate is 2+8i (YES, YOU ARE CORRECT) (or in fact it could also be -2-8i, but THAT would be a negative conjugate)
anonymous
  • anonymous
Coolio. This one is giving me trouble as well. Simplify the expression |dw:1449680910058:dw|
anonymous
  • anonymous
Answer Choices: (Just for reference) A. \[\frac{ 12-9i }{ 25 }\] B. \[\frac{ 12-9i }{ 7 }\] C. \[\frac{ 9+12i }{ 25 }\] D. \[\frac{ 9+12i }{ 7 }\]
anonymous
  • anonymous
I chose C.
SolomonZelman
  • SolomonZelman
You are probably aware of these two facts: \(\large\color{#000000 }{ \displaystyle (a-b)(a+b)=a^2-b^2 }\) and, \(\large\color{#000000 }{ \displaystyle i^2=-1 }\) So, the conjugate is there to eliminate the i. So if you have \(\large\color{#000000 }{ \displaystyle (2i+4) }\), then as you multiply that times the conjugate (in this case (2i-4), you get: \(\large\color{#000000 }{ \displaystyle(2i+4)(2i-4)=(2i)^2-4^2=2^2\cdot i^2-16 }\) \(\large\color{#000000 }{ \displaystyle=4\cdot (-1)-16 =-4-16=-20 }\) But, why do we need the CONJUGATE tho, if we can't just randomly multiply every imaginary number times its conjugate (because you are unjustly changing the value). Suppose you had the following problem. \(\large\color{#000000 }{ \displaystyle \frac{i+1}{2i+4} }\) and you want to rationalize the denominator (remove any roots from the denominator) Then, you would multiply top and bottom by the conjugate of 2i+4 (which is 2i-4) \(\large\color{#000000 }{ \displaystyle \frac{(i+1)\color{red}{\times (2i-4)}}{(2i+4)\color{red}{\times (2i-4)}} }\) You already know that on the bottom you will get a -20. \(\large\color{#000000 }{ \displaystyle \frac{(i+1)\color{red}{\times (2i-4)}}{-20} }\) and as we simplify the to we get: \(\large\color{#000000 }{ \displaystyle (i+1)\times (2i-4) }\) \(\large\color{#000000 }{ \displaystyle (i+1)\times 2i=2i^2+2i=2(-1)+2i=-2+2i }\) \(\large\color{#000000 }{ \displaystyle (i+1)\times 1=i+1 }\) add the results, you get: \(\large\color{#000000 }{ =-2+2i +i+1=-1+3i \quad or,~=3i-1 }\) So \(\large\color{#000000 }{ \displaystyle (i+1)\times (2i-4)=3i-1 }\) And thus you get: \(\color{#000000 }{ \displaystyle\frac{i+1}{2i+4}= \frac{(i+1)\color{red}{\times (2i-4)}}{(2i+4)\color{red}{\times (2i-4)}}=\frac{3i-1}{-20} }\) which can also be written as (1-3i)/20.
SolomonZelman
  • SolomonZelman
In case of you problem, you have to add like terms in the denominator first, and then use my example to rationalize the denominator.
SolomonZelman
  • SolomonZelman
One note for the NUMERATOR, though, \(\large\color{#000000 }{ \displaystyle \sqrt{-9}=\sqrt{(-1)}\cdot \sqrt{9}=i\times 3=3i }\)
anonymous
  • anonymous
That's a lot of information. Thanks.
SolomonZelman
  • SolomonZelman
I will provide the steps, just in case: { Step 1 } Your denominator is (3-2i)+(1+5i). Simplify the denominator. { Step 2 } Multiply top and bottom times the conjugate of the new denominator.
SolomonZelman
  • SolomonZelman
Note, for step 2, apply, (a-b)(a+b)=a²-b² just as I did, and then simplify. Also don't forget that i²=-1
anonymous
  • anonymous
I got this answer this time: \[\frac{ 9+12i }{ 7 }\]
anonymous
  • anonymous
Is this correct? @SolomonZelman
SolomonZelman
  • SolomonZelman
you made a little error (4+3i)(4-3i) = is NOT 16-9 RATHER, (4+3i)(4-3i) = is going to be = 16 + 9
SolomonZelman
  • SolomonZelman
because 3i × (-3i)= -3i² = -3(-1) = 3.
anonymous
  • anonymous
Yay! I got a 100%! Thank you so much for your time, patience, and help! @SolomonZelman

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