$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 6 &,& 8~) 
% (c,d)
&&(~ 8 &,& 12~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-8}{8-6}\implies \cfrac{4}{2}\implies 2
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-8=2(x-6)
\\\\\\
y-8=2x-12\implies y=2x-4$

7 0

4 years ago

What is the smallest number with the factors 1,2,3,4, and 5? What is the special name for it?

klasskru [66]

To find LCM
factor and eliminate ones already represented
1=1
2=2
3=3
4=2*2
5=5

1*2*2*3*5=60

7 0

3 years ago

Prove that x 2 + y2 - 1 is irreducible in q[ x, y].

weqwewe [10]

Suppose that the polynomial x^2+y^2-1 is not irreducible.
It means that there exists two fractions such that :
$(\frac{p}{q})^2+( \frac{s}{t} )^2 -1=0\\\text{rearranging the above formula we get:}\\p^2t^2=q^2(t^2-s^2)\\\text{since q and p are primes each other, we deduce that }\\\text{q divide t meaning that we can write }t=a\times t\text{ for an integer a}\\\text{rearranging again the formula and using our finding we get:}\\q^2s^2=a^2q^2(q^2-p^2)\\\text{Same argument as above, we deduce that q divide s.}$
We found that q divide t and q divide s, but s and t are supposed to be prime each other. It is a contradiction.
The polynomial x^2+y^2-1 is then irreducible in q[x,y]

6 0

4 years ago

A triangle has an area of 125 square inches. The height of the triangle is 8 inches. Which equation

Free_Kalibri [48]

Answer:

(125*2)/8=b

Step-by-step explanation:

The way to find the area of a triangle is to multiply height and width then divide by two and 125 is the area. Then that means 125=250/2.

Since the height is 8 we can make the equation (125*2)/8.

Hopefully that was helpful and made sense.

5 0

3 years ago