5.10 Segment Honors Project
By:Rashone Mills
Task 5
Yes, Honor standards are really necessary. I feel like it sets a standard for students that feel qualified to extend their knowledge to another level. The lessons in this honor standards are used in the real world in many occasions for example, the rational expressions are used daily. The average rate of change, the difference in cubes and much more. Yes, I feel that it is valuable enough to be funded it deserves to be funded because it helps out the economy when kids are gaining more knowledge and are getting smarter by the day.
Task 2
1.) (x+2)^6
By using the binomial theorem we will get: x^6+12x^5+60x^4+160x^3+240x^2+192x+64
2.) (x-4)^4
By using the binomial theorem we will get:
x^4-16x^3+96x^2-256x+256
3.) (2x+3)^5
By using the binomial theorem we will get:
32x^5+240x^4+720x^3+1080x^2+810x+243
4.) (2x-3y)^4
By using the binomial theorem we will get:
16x^4-96x3y+2162y2-216xy3+81y4
5.) Expansion of (3a+4b)^8
I picked ab^7 and there's more where the exponents add to 8.
a2b3, a5b3, ab8, b8, a4b4, a8, ab7, a6b5
Task 3
1.) f(x)=x^4+21x^2-100
roots for the operation is: (x^2+25) (x^2-4)
2.) f(x)=x^3-5x^2-25x+125
roots for the operation is: (x-5)^2 (x+5)
3.) Part1: The polynomial has even and odd multiplicities. Because the function crosses the x-axis at the root, and the even multiplicity the function touches the graph at the root. The polynomial zero's are -5, -2, 3, 7
Part2.: The degree of the polynomial is 7. the possible factored form is (x^2+25) (x-2) (x+2) (x-5) 2(x+5) without plotting any points other than intercepts, draw a possible graph of the following polynomial: f(x)= (x+8) 3(x+6) 2(x+2) (x-1) 3(x-3) 4(x-6).
Task 1
1.)Factor x^2+64 check your work
X^2-(-64)
(x+8i) (x-8i)
Check your work: (x+8i) (x-8I)= x^2-8i+8i-64i^2=x^2-64(-1)=x^2+64
2.)Factor 16x^2+49
16x^2-(-49)
(4x+7i) (4x-7i)
Check your work: (4x+7i) (4x-7i)= 16x^2 – 7i + 7i – 49i^2 = 16x^2-49(-1)= 16x^2+49
3.)Product of (x+9i)^2
X^2+2(x*9i)+(9i)^2
X^2+18ix + 81i^2
X^2+18ix-81
4.)Product of (x-2i)2
x^2+2(x*-2i)+(-2i)^2
x^2+-4ix+4i^2
x^2-4ix-4
5.) (x+(3+5i))^2
x^2+2(x*(3+5i))+(3+5i)^2
x^2+2(3x+5ix)+(6+20i+25i^2)
x^2+6x+10ix+6+20i-25
X^2+6x+9
Thank you!
Task 4
1.) X/x+3 + x + 2/ x+5= The closure for this following operation is: 2(x^2+5x+3) (x+3) (x+5)
2.) x+4/ x^2 + 5x + 6*x+3/ x^2-16= The closure for the following operation is: 1/(x+2)(x-4)
3.) 2/x^2 - 9 - 3x/ x^2 - 5x + 6= The closure for this following operation is: -3x^3+2x^2-17x+12/ (x^2-9) (x^2-5x+6)
4.) x+4/x^2-5x+6/x^2-16/x+3 the closure for this following operation is: (x+4)^2 (x-4)/ (x-2) (x-3) (x+3)
Compare&Contrast: The divisions are very similar. But the divisions of integers compared to the ones of rational expression is much easier in my opinion.
Rashone Mills
5.10 Segement Honors Project
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5.10 Segment Honors Project
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