Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.
y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).
Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,
-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.
The equation of parabola y4 is y+4 = (1/4)x^2
Or you could elim. the fraction and write the eqn as 4y+16=x^2, or
4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.
Answer: It is B)
Step-by-step explanation: It was 8.66 something and that rounded to the nearest tenth is 8.7
X=m/5-p/5 i tried my best to answer this problem
With the concept of first in, first out method, then we
can use the formula below to solve for the number of equivalent units of
production for that period.
number of equivalent units of production
= Total number of units completed during that period (A) –
Number of units completed in process at the beginning of the period (B) +
Number of units completed at the end of the period (C)
= A – B + C
We know that,
A = 9000 units
So we solve for B and C.
B is 60% of the 500 units, therefore:
B = 0.60 * 500 = 300
C is 30% of the 600 units, therefore:
C = 0.30 * 600 = 180
Substituting the values into the equation:
number of equivalent units of production = 9000 – 300 + 180
number of equivalent units of production = 8880 units
Answer:
A. 8880
Answer:
ST = 12 units
Step-by-step explanation:
Δ RAS and Δ TES are similar triangles, thus the ratios of corresponding sides are equal, that is, noting that SR = 21 - ST, thus
= 
substitute values
= 
( cross- multiply )
6ST = 8(21 - ST)
6ST = 168 - 8ST ( add 8ST to both sides )
14ST = 168 ( divide both sides by 14 )
ST = 12
