To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
Answer:
B. Place the compass at P and draw an arc of length OP across the circle.
Below is the diagram for help.
Step-by-step explanation:
We are given with an already constructed circle of diameter
and
as radius.
To draw a regular hexagon(six-sided), we need to make first cut on the circle of an arc equal to radius of the circle taking any point of the diameter as center
So here, We will consider
as a center and take an arc of length
and make a cut on the circle.
This would be the first step and James need to repeat the same steps by considering the newly formed cut as center each time.

Actually Welcome to the Concept of the Trigonometry.
here, we use the Linear pair property of the adjacent angles.
We know that, all the adjacent angles in a linear pair add to get 180°
so we get as,
=> (n+6) +90°+(2n+3) =180°
=> (n+6) +(2n+3) =180-90
=> 3n+9 = 90
=> 3n= 90-9
=> 3n = 81
hence, n = 81/3
=> n = 27°
thus the value of n is 27° .
Answer:
x = 8
Step-by-step explanation:
since it's a collinerar
or, AB+BC=AC
or, 3x+5x+10 = 74
or, 8x = 74-10
or, 8x = 64
or, x = 64/8
x = 8
Answer:
Spencer and Abigail are correct
Lauren is incorrect
Step-by-step explanation:
Spencer and Abigail are correct
Slope = change in y ÷ change in x
Or 
Let
= (3, -1)
Let
= (5, 4)

This is Spencer's method
Let
= (5, 4)
Let
= (3, -1)

This is Abigail's method
It doesn't matter which point you label as point 1 and point 2, as long as you carry out the slope calculation correctly.
Lauren's calculation is wrong as she calculated her slope as:
where it should have been 