Answer:
cot∅ = (-2√30)/7.
Step-by-step explanation:
Given the value of csc∅ = -13/7 and ∅ is in quad III.
We know y = r sin∅ and r > 0. So csc∅ = r/y = -13/7 = 13/(-7).
It means y = -7, r = 13.
We know x² + y² = r².
x² = r² - y²
x² = (13)² - (-7)² = 169 - 49 = 120.
x = √120 = 2√30.
we know cot∅ = x/y = (2√30)/(-7) = (-2√30)/7.
Hence, cot∅ = (-2√30)/7.
Answer:
(1, 25)
Step-by-step explanation:
Convert to vertex form so that the equation is (x - 1)^2 + 25
If 1 is <em>h </em>and 25 is <em>k</em>, (<em>h, k</em>), the result would be (1, 25)
Answer:
y approaches negative infinty
Answer:
B should be 3 inches long because the area of the square above is 48 inches squared, so that long side is probably 8. B is 3 inches long. As for A, this should be simple. since we know the longer side of the room has the area of 27, we can tell the length(longest side) is 9. 14 minus 9 is equal to 4. A's width is 4 inches long.
Step-by-step explanation:
for the rest, since we know one of the widths is 6 inches, we know that the length of the room with an area of 32 inches is 8. therefore, since 8 times 4 equals the number 32, C is 4 inches long. back to B. since we know that c is 4 inches long and that there is a length of 11 inches, we can tell the longest side of room A is 7. 4 Times 7 equals 28, so B is 28 inches squared.
Note that the 2nd equation can be re-written as y=8x-10.
According to the second equation, y=x^2+12x+30.
Equate these two equations to eliminate y:
8x-10 = x^2+12x+30
Group all terms together on the right side. To do this, add -8x+10 to both sides. Then 0 = x^2 +4x +40. You must now solve this quadratic equation for x, if possible. I found that this equation has NO REAL SOLUTIONS, so we must conclude that the given system of equations has NO REAL SOLUTIONS.
If you have a graphing calculator, please graph 8x-10 and x^2+12x+30 on the same screen. You will see two separate graphs that do NOT intersect. This is another way in which to see / conclude that there is NO REAL SOLUTION to this system of equations.
