Answer:
P ≈ 100
Step-by-step explanation:
260 = P(1+0.054/365)^365(14)
260 = P(1.000147945)^5110
260= P(2.129621112)
P= 122.087444818
P ≈ 100
Option a, c,d are correct.
step-by-step explanation:
from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have
tz=sz(given)
bz=zb(common)
therefore, by rhs rule,δtzb ≅δszb
by cpctc, sz≅tz
also, from δctz and δasz,
tz=sz(given)
∠tcz=∠saz(90°)
by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz
also,from δasz and δzsb,
zs=sz(common)
∠zbs=∠saz=90°
by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb
hence, option a, c,d are correct.
Answer:
The fourth vertex is D(8, 0)
Step-by-step explanation:
Let A(0, 0), B(2, 4), and C(10, 4) be the three vertices of a parallelogram ABCD and let its fourth vertex be D(a, b).
Join AC and BD. Let AC and BD intersect at point O.
- We have known that the diagonals of a parallelogram bisect each other.
So, O would be the midpoint of AC as well as that of BD.
The midpoint of AC is:
The midpoint of BD is:
so
∵ 
∵ 
Hence, the fourth vertex is D(8, 0)
Hey there! :)
Answer:
Choice 4: π/4, 3π4, 7π/4
Step-by-step explanation:
For the values of θ to have the same reference angles, the denominators of the radians must be the same. Therefore:
Choice 1: π/6, π/3, 5π/6 - incorrect. π/3 will have a different reference angle as it has a denominator of 3.
Choice 2: π/3, 5π/6, 4π/3 - incorrect. 5π/6 will have a different reference angle as it has a denominator of 6.
Choice 3: π/2, 5π/4, 7π/4 - incorrect. π/2 will have a different reference angle.
Choice 4: π/4, 3π4, 7π/4 - correct. All of these radians contain the same denominator, and will each have the same reference angles of π/4.
The answer is Angle BCA. I’m pretty sure
