For this case we have the following type of equations:
Quadratic equation:
Linear equation:
We observe that when equating the equations we have:
Rewriting we have:
We obtain a polynomial of second degree, therefore, the maximum number of solutions that we can obtain is 2.
Answer:
The greatest number of possible solutions to this system is:
c.2
Given: In the given figure, there are two equilateral triangles having side 50 yards each and two sectors of radius (r) = 50 yards each with the sector angle θ = 120°
To Find: The length of the park's boundary to the nearest yard.
Calculation:
The length of the park's boundary (P) = 2× side of equilateral triangle + 2 × length of the arc
or, (P) = 2× 50 yards + 2× (2πr) ( θ ÷360°)
or, (P) = 2× 50 yards + 2× (2×3.14× 50 yards) ( 120° ÷360°)
or, (P) = 100 yards + 2× (2×3.14× 50 yards) ( 120° ÷360°)
or, (P) = 100 yards + 209.33 yards
or, (P) = 309.33 yards ≈309 yards
Hence, the option D:309 yards is the correct option.
(1) -(5y - 2)= -5y+2
(2) -5 (3n + 1)= -15n - 5
(3) too long to do
Answer:
it is 1
Step-by-step explanation:
9514 1404 393
Answer:
31.243 units
Step-by-step explanation:
The mnemonic SOH CAH TOA reminds you of the relationships between sides and angles in a right triangle. Using the attached figure, it is convenient to find the length of BE as an intermediate step in the solution.
Sin = Opposite/Hypotenuse
sin(30°) = BE/100
BE = 100·sin(30°)
Then ...
Tan = Opposite/Adjacent
tan(58°) = BE/x
x = BE/tan(58°) = 100·sin(30°)/tan(58°)
x ≈ 31.243 . . . . units
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<em>Comment on the figure</em>
The intermediate problem in creating the figure was to locate point D. That was accomplished by locating point C on a line at an angle of 58° CCW from the horizontal, using point B as a center. Then D is the intersection of BC with the x-axis. BE is drawn perpendicular to the x-axis.
