Answer:
Yes because
Step-by-step explanation:
Is 3z-1 in standard form?
Yes becausestandard form of a polynomial, the value in the power of the variable such as z in the equation reduces from left to right. Therefore, 3z - 1 is in standard form since the z term comes to the left of the constant term
Another word for elasticity is also called D: responsiveness
First, some housekeeping:
cos = 12/13 is incomplete; "cos" must have an argument (input).
cos x = 12/13 is fine; here "cos" has the argument (input) x.
Given that cos x = 12/13, find sin x. To do this, we'll need to find the length of the opposite side, given that the hypo length is 13 and the adj. side length is 12.
12^2 + opp^2 = 13^2, or opp^2 = 169-144 = 25.
Then the opp side could be either 5 or -5. Let's assume that it's +5, and that angle x is in the first quadrant.
Then sin x = opp / hyp = 5/13 (answer)
cos 2 is an entirely different kind of problem. Here you are told what the argument (input) to the cosine function is (it is 2, which here means 2 radians).
Using a calculator: cos 2 = -0.416. Note that the angle 2 rad is in QII, which is why the "adjacent side" is negative and also why the cos of 2 is negative.
Answer: BC = 5.83
Step-by-step explanation:
Luckily, the triangle is placed on the graph nicely so we can count the legs of the triangle:
AB = 5
AC = 3
BC = ?
To find BC, we can simply use the Pythagorean Theorem:
5^2 + 3^2 = c^2
25 + 9 = c^2
34 = c^2
Now square root to find c, or BC.
c = 5.83 (rounded by nearest hundredth)
Answer:
5*3=15
Step-by-step explanation:
