A six-sided number cube is rolled twice. Which expression can be used to find P(5, then3)?.
1 answer:
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Evaluate each of the expressions for the given values of the variables and see if the condition is met.
8000*4 +5500*6 = 65000 ≤ 75000 . . . true
52*4 +38*6 = 436 ≥ 320 . . . . . . . . . . . . . true
Yes, the company can meet both its expectations by building 4 cars and 6 motorbikes.
Of course, another way to work this is to draw the graph and see if the offered point is in the feasible region. It is.
8000*4 +5500*6 = 65000 ≤ 75000 . . . true
52*4 +38*6 = 436 ≥ 320 . . . . . . . . . . . . . true
Yes, the company can meet both its expectations by building 4 cars and 6 motorbikes.
Of course, another way to work this is to draw the graph and see if the offered point is in the feasible region. It is.
Answer:
Step-by-step explanation:
Here you are require to find the equation of the hyperbola given that the center (h,k), the coordinates of the vertices and those of the co-vertices can be determined from the diagram given
The sharp turning points of the curves give the vertices at (-4,2) and (6,2)
Joining the vertices with a straight line will form the transverse axis with length 2a .
To find the length of the transverse axis 2a will be ; 6--4=10. 2a=10 hence a=10/2 =5
a=5
Find the center of the hyperbola at (h,k) by finding the intersecting point of the diagonals of the rectangle in the diagram
The center identified will be (h,k) = (1,2)
To find the length of the conjugate axis 2b will be ; the length between points (1,4) and (1,0) which are the coordinates of the co-vertices in the hyperbola. 2b= 4-0=4 , b=4/2 = 2
b=2
The standard equation of the hyperbola with center (h,k) is written as ;
where (h,k) is center of hyperbola, (h±a,k) is coordinate of the vertices and (h,k±b) are coordinates of co-vertices.
Substitute values of a, b, h, and k in equation as