A telephone pole is 20 feet tall. If the shadow of the telephone pole is 65 and
1 answer:
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Answer:
Step-by-step explanation:
The question to be solved is the following :
Suppose that a and b are any n-vectors. Show that we can always find a scalar γ so that (a − γb) ⊥ b, and that γ is unique if . Recall that given two vectors a,b a⊥ b if and only if
where
is the dot product defined in
. Suposse that
. We want to find γ such that
. Given that the dot product can be distributed and that it is linear, the following equation is obtained
Recall that are both real numbers, so by solving the value of γ, we get that
By construction, this γ is unique if , since if there was a
such that
, then
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.
