So slope intercept form is y=mx+b
m=slope b=y intercept
convert current equation to slpe intercept
y-4=x-8
add 4 to both sides
y=x-4
tada
The dimensions of the prism can be 2x, 2x+3 and x+6.
We first factor out the GCF of the trinomial. The GCF of the coefficients is 2. Each term has an x in common as well, so the GCF is 2x.
Factoring out the 2x, we have
2x(2x²+15x+18).
To factor the remaining trinomial, we find factors of 2*18=36 that sum to 15. 12*3 = 36 and 12+3 = 15. We split up 15x into 12x and 3x:
2x(2x²+12x+3x+18)
Now we group together the first two terms in parentheses and the last two:
2x((2x²+12x)+(3x+18))
Factor out the GCF of the first group:
2x(2x(x+6)+(3x+18))
Factor out the GCF of the second group:
2x(2x(x+6)+3(x+6))
Factoring out what these have in common,
2x(x+6)(2x+3)
each term is negative and 1/4 of previous term so the nth term is the n-1 term times -1/4 so f(n)= -1/4 f(n-1)
We have 3⁴ = 81, so we can factorize this as a difference of squares twice:
Depending on the precise definition of "completely" in this context, you can go a bit further and factorize as yet another difference of squares:
And if you're working over the field of complex numbers, you can go even further. For instance,
But I think you'd be fine stopping at the first result,
Answer:
a. 25.98i - 15j mi/h
b. 1.71i + 4.7j mi/h
c. 27.69i -10.3j mi/h
Step-by-step explanation:
a. Identify the ship's vector
Since the ship moves through water at 30 miles per hour at an angle of 30° south of east, which is in the fourth quadrant. So, the x-component of the ship's velocity is v₁ = 30cos30° = 25.98 mi/h and the y-component of the ship's velocity is v₂ = -30sin30° = -15 mi/h
Thus the ship's velocity vector as ship moves through the water v = v₁i + v₂j = 25.98i + (-15)j = 25.98i - 15j mi/h
b. Identify the water current's vector
Also, since the water is moving at 5 miles per hour at an angle of 20° south of east, this implies that it is moving at an angle 90° - 20° = 70° east of north, which is in the first quadrant. So, the x-component of the water's velocity is v₃ = 5cos70° = 1.71 mi/h and the y-component of the water's velocity is v₄ = 5sin70° = 4.7 mi/h
Thus the ship's velocity vector v' = v₃i + v₄j = 1.71i + 4.7j mi/h
c. Identify the vector representing the ship's actual motion.
The velocity vector representing the ship's actual motion is V = velocity vector of ship as ship moves through water + velocity vector of water current.
V = v + v'
= 25.98i - 15j mi/h + 1.71i + 4.7j mi/h
= (25.98i + 1.71i + 4.7j - 15j )mi/h
= 27.68i -10.3j mi/h