Find the distance between the points (4,10) and (0,0) the answer must be a whole number
1 answer:
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Answer:
Lines a and b are parallel
Lines a and c are perpendicular
Lines d and c are perpendicular
Step-by-step explanation:
we know that
The formula to calculate the slope between two points is equal to
Part 1) Find the slope of Line a
we have the points
(-3,4) and (3,6)
substitute in the formula
simplify
Part 2) Find the slope of Line b
we have the points
(-10,-3) and (-8,3)
substitute in the formula
Part 3) Find the slope of Line c
we have the points
(0,5) and (3,-4)
substitute in the formula
Part 4) Find the slope of Line d
we have the points
(4,-7) and (13,-4)
substitute in the formula
simplify
Part 5) Compare the slopes
Remember that
If two lines are parallel then their slopes are the same
If two lines are perpendicular then their slopes are opposite reciprocal
we have
therefore
Lines a and b are parallel (slopes are equal)
Lines a and c are perpendicular (slopes are opposite reciprocal)
Lines d and c are perpendicular (slopes are opposite reciprocal)
Answer:
3) Cubic polynomial with four terms.
4) Linear polynomial with two terms.
5) The zeros are: (1-sqrt(5))/2=-0.618, 1, and (1+sqrt(5))/2=1.618
The y-intercept is y=1
Please, see the attached graph.
6) The zeros are: -1.272, 1.272
The y-intercept is y=1
Please, see the attached graph.
7) Please, see the attached files.
8) Please, see the attached files.
9) Please, see the attached files.
10) Please, see the attached files.
Step-by-step explanation:
3) Degree is the maximun exponent that the variable "x" has, in this case 3, then this is a cubic polynomial, and it has four terms (6x^3, -7x^2, -10x, and -8).
4) Degree is the maximun exponent that the variable "x" has, in this case 1 (x=x^1), then this is a linear polynomial, and it has two terms (-10x, and 10).
5) f(x)=x^3-2x^2+1
Zeros:
f(x)=0→x^3-2x^2+1=0
Factoring:
x^3-2x^2+0x+1
! 1 -2 0 1
<u>1 !____1____-1_-1</u>
1 -1 -1 0
1x^2-1x -1 = x^2-x-1
x^3-2x^2+1=0→(x-1)(x^2-x-1)=0
The zeros are:
x-1=0→x-1+1=0+1→x1=1
x^2-x-1=0
ax^2+bx+c=0; a=1, b=-1, c=-1
x=(-b+-sqrt(b^2-4ac))/(2a)
x=(-(-1)+-sqrt((-1)^2-4(1)(-1))/(2(1))
x=(1+-sqrt(1+4))/2
x2=(1+-sqrt(5))/2=(1+2.236067978)/2=3.236067978/2=1.618033989=1.618
x=(1-sqrt(5))/2=(1-2.236067978)/2=-1.236067978/2=-0.618033989=-0.618
x3=(1+sqrt(5))/2
The zeros are: (1-sqrt(5))/2=-0.618, 1, and (1+sqrt(5))/2=1.618
The y-intercept is:
x=0→f(0)=(0)^3-2(0)^2+1→f(0)=0-2(0)+1→f(0)=0-0+1→f(0)=1
The y-intercept is y=1.
6) f(x)=-x^4+x^2+1
Zeros:
f(x)=0→-x^4+x^2+1=0
Factoring: Multiplyng both sides of the equation by -1:
(-1)(-x^4+x^2+1)=(-1)(0)
x^4-x^2-1=0
(x^2)^2-(x^2)-1=0
Changing x^2 by t:
t^2-t-1=0
at^2+bt+c=0; a=1, b=-1, c=-1
t=(-b+-sqrt(b^2-4ac))/(2a)
t=(-(-1)+-sqrt((-1)^2-4(1)(-1))/(2(1))
t=(1+-sqrt(1+4))/2
t1=(1+-sqrt(5))/2=(1+2.236067978)/2=3.236067978/2=1.618033989=1.618
t2=(1-sqrt(5))/2=(1-2.236067978)/2=-1.236067978/2=-0.618033989=-0.618
t^2-t+1=0→(t-1.618)(t-(-0.618))=0→(t-1.618)(t+0.618)=0
and t=x^2, then:
(x^2-1.618)(x^2+0.618)=0
Factoring the first parentheses using a^2-b^2=(a+b)(a-b), with:
a^2=x^2→sqrt(a^2)=sqrt(x^2)→a=x
b^2=1.618→sqrt(b^2)=sqrt(1.618)→b=1.272
(x^2-1.618)(x^2+0.618)=0→(x+1.272)(x-1.272)(x^2+0.618)=0
The zeros are:
x+1.272=0→x+1.272-1.272=0-1.272→x1=-1.272
x-1.272=0→x-1.272+1.272=0+1.272→x2=1.272
The zeros are: -1.272, and 1.272
The y-intercept is:
x=0→f(0)=-(0)^4+(0)^2+1→f(0)=-0+0+1→f(0)=1
The y-intercept is y=1.
