Think of the space created with the diagonal line like a triangle. There are three sides. the longest side, the diagonal line is 32 inches. one of the other sides is 15.7 inches. Using Pythagoras theorem, we know that a^2 + b^2 = c^2, where c is the longest side of the triangle (32inches) and either a or b is 15.7inches. let's say that a is 15.7 inches. if we substitute these values we just assigned to a and c into Pythagoras' theorem above, we get 15.7^2 +b^2 = 32^2. we can simplify this to 246.49 + b^2 = 1024. We then subtract 246.49 from both sides of the equation to get b^2 = 777.51. we then square root both sides of the equation to get b = 27.88 (rounded to 2 decimal places). because b is the width of the TV, the width of the TV is 27.88 inches.
4y - x = 5 + 2y ..... (1)
3x + 7y = 24 ..... (2)
by grouping like terms in (1)
4y - x = 5 + 2y
4y - 2y - x = 5
<span>-x + 2y = 5 </span> ..... (1a)
by multiplying (1a) through by -3
(-3)(-x) + 2(-3)y = 5(-3)
3x - 6y = -15 ..... (1b)
by subtracting 1a from 2
3x -3x + 7y - (-6y) = 24 - (-15)
13y = 39
⇒ y = 3
by substituting y=3 into (2)
3x + 7(3) = 24
3x = 24 - 21
3x = 3
⇒ x = 1
∴ solution to the system is x=1 when y = 3
Answer:
Step-by-step explanation:
When you reflect over the line y=-x, you flip the x and y coordinate and multiply them by negative one, so for example (2,3) would become (-3,-2) or (-1,7) would be (-7,1). Using this rule you can find the new point for each point in the figure. The point lying on (-2,1) would be reflected to (-1,2). The point on (-4,1) would move to (-1, 4). The point on (-4,-1) would be moved to (1,4). The point lying on (-3,-1) would be reflected onto (1,3). The point on (-3,-2) would be on (2,3). Lastly, the point on (-2,-2) would be reflected to (2,2). i hope this helped lol. :)
Answer: 0, π, 2π
Nevertheless, that is not an option. I see two possibilities: 1) the options are misswirtten, 2) the domain is not well defined.
If the domain were 0 ≤ θ < 2π, then 2π were excluded of the domain ant the answer would be 0, π.
Explanation:
1) The first solution, θ = 0 is trivial:
sin (0) - tan (0) = 0
0 - 0 = 0
2) For other solutions, work the expression:
sin(θ) + tan (-θ) = 0 ← given
sin (θ) - tan(θ) = 0 ← tan (-θ) = tan(θ)
sin(θ) - sin (θ) / cos(θ) = 0 ← tan(θ) = sin(θ) / cos(θ)
sin (θ) [1 - 1/cos(θ)] = 0 ← common factor sin(θ)
⇒ Any of the two factors can be 0
⇒ sin (θ) = 0 or (1 - 1 / cos(θ) = 0,
sin(θ) = 0 ⇒ θ = 0, π, 2π
1 - 1/cos(θ) = 0 ⇒ 1/cos(θ) = 1 ⇒ cos(θ) = 1 ⇒ θ = 0, 2π
⇒ Solutions are 0, π, and 2π
In fact if you test with any of those values the equation is checked. The only way to exclude one of those solutions is changing the domain.
