It is a bit tedious to write 6 equations, but it is a straightforward process to substitute the given point values into the form provided.
For segment ab. (x1, y1) = (1, 1); (x2, y2) = (3, 4).
... x = 1 + t(3-1)
... y = 1 + t(4-1)
ab = {x=1+2t, y=1+3t}
For segment bc. (x1, y1) = (3, 4); (x2, y2) = (1, 7).
... x = 3 + t(1-3)
... y = 4 + t(7-4)
bc = {x=3-2t, y=4+3t}
For segment ca. (x1, y1) = (1, 7); (x2, y2) = (1, 1).
... x = 1 + t(1-1)
... y = 7 + t(1-7)
ca = {x=1, y=7-6t}
Answer:
4^8
Step-by-step explanation:
If the second 4 is an exponent, as in (4^2)^4, then multiply the exponents.
(4^2)^4 = 4^(2 * 4) = 4^8
In
order to solve for a nth term in an arithmetic sequence, we use the formula
written as:<span>
an = a1 + (n-1)d
where an is the nth term, a1 is the first value
in the sequence, n is the term position and d is the common difference.
</span><span>THIRD
</span><span>A3=4+(3-1)(-5)
A3 = -6
A(3)=-2(3-1)(-5)
A3 = 20
</span><span>
FIFTH
</span>A5=4+(5-1)(-5)
A5 = -16
A(5)=-2(5-1)(-5)
A5 = 40<span>
TENTH
</span>A10=4+(10-1)(-5)
A10 = -41
A(10)=-2(10-1)(-5)
A10 = 90