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crimeas [40]
3 years ago
5

If the points A, B and C have the coordinates A(5, 2), B(2, −3) and C(−8, 3), show that the triangle ABC is a right-angled trian

gle.
Mathematics
1 answer:
bixtya [17]3 years ago
6 0

Answer:

See below.

Step-by-step explanation:

Triangle ABC will be right-angled if 2 of its sides are perpendicular to each other. When sides are perpendicular the product of their slopes = -1.

Slope  of side AB = (-3-2)/(2-5) = -5/-3 = 5/3

Slope of side BC = (3- -3) / -8-2) = 6/-10 = -3/5

The product of the slopes = 5/3 * -3/5 = -15/15 = -1.

So AB is perpendicular to BC  and triangle ABC is right-angled .

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High school math. Please answer everything in picture. Thank you................................................................
vaieri [72.5K]

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.

3 0
3 years ago
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