The tens digit of a two-digit number is twice the units digit. If the digits are reversed, the new number is 36 less than the or
1 answer:
Try this solution:
let the unknown two-digits number be 'xy'.
Then it is possible to write any two-digit number as 10x+y, where x - the tens digit and y - the unit digit.
Using the rule described above the phrase 'the tens digit of a number is twice the units digit' may be written as x=2y.
Using the same rule the phrase 'if the digits are reversed, the new number is 36 is less than the original number' may be written as (10x+y)-(10y+x)=36.
These two equations may be resolved as system of two equations:
Check: the original number is 84, the new number is 48; difference is 84-48=36.
answer: 84
let the unknown two-digits number be 'xy'.
Then it is possible to write any two-digit number as 10x+y, where x - the tens digit and y - the unit digit.
Using the rule described above the phrase 'the tens digit of a number is twice the units digit' may be written as x=2y.
Using the same rule the phrase 'if the digits are reversed, the new number is 36 is less than the original number' may be written as (10x+y)-(10y+x)=36.
These two equations may be resolved as system of two equations:
Check: the original number is 84, the new number is 48; difference is 84-48=36.
answer: 84
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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
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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.
