For perpendicular lines m2 = -1/m1, i.e. the product of the two slopes equals -1.
x - 5y = 3
5y = x + 3
y = 1/5 x + 3/5 => slope = 1/5
m2 = -1/(1/5) = -5
Equation of a line containing point (-4, 6) with a slope of -5 is
y - 6 = -5(x - (-4))
y - 6 = -5(x + 4)
y - 6 = -5x - 20
5x + y = -20 + 6
5x + y = -14
This object has rotational symmetry because if you made a partial turn it would still be the same all around, it doesn't have line symmetry tho because you couldn't draw a line through it and have it equal on both sides
Answer:
The relationships, properties, and theorems will be easier to understand when ... Just like you don't want to do a geometry problem without all the given ... In case you want to get a head start… here are arguably some of the most important theorems for triangles: ... In the figure below, both congruent angles have one arch.
Step-by-step explanation:
Answer:
Their are caculators online search the type of math it is then put caculator for example exponents caculator
Step-by-step explanation:
Answer:
(1,2021)
Step-by-step explanation:
P and q can vary subject to their sum being 2020.
Consider one parabola with p1 and q1 and another with p2 and q2.
y1=(x1)^2+(p1)(x1)+(q1)
y1=(x2)^2+(p2)(x2)+(q2)
At their intersection, the x and y coordinates are the same.
y1=y2=y
x1=x2=x
x^2+(p1)x+(q1)=x^2+(p2)x+(q2)
Solve for x
x(p1-p2)=q2-q1
x=(q2-q1)/(p1-p2)
Use the constraint that p+q=2020 to eliminate p1 and p2.
p1=2020-q1
p2=2020-q2
x=(q2-q1)/(2020-q1-2020+q2)
x=(q2-q1)/(q2-q1)
x=1
Substitute in the equation for y.
y=1^2+p(1)+q
y=2021