Answer:
∠JKL = 38°
Step-by-step explanation:
PQRS, JQK and LRK are straight lines
Let's take the straight lines in the diagrams one after the other to find what they consist.
The related diagram can be found at brainly (question ID: 18713345)
Find attached the diagram used for solving the question.
For straight line PQRS,
2x°+y°+x°+2y° = 180°
(Sum of angles on a Straight line = 180°)
Collect like terms
3x° + 3y° = 180°
Also straight line PQRS = straight line PQR + straight line SRQ
For straight line PQR,
2y + x + ∠RQM = 180° ....equation 1
For straight line SRQ,
2x + y + ∠MRQ = 180° ....equation 2
Straight line PQRS = addition of equation 1 and 2
By collecting like times
3x +3y + ∠RQM + ∠MRQ = 360°....equation 3
Given ∠QMR = 33°
∠RQM + ∠MRQ + ∠QMR = 180° (sum of angles in a triangle)
∠RQM + ∠MRQ + 33° = 180°
∠RQM + ∠MRQ = 180-33
∠RQM + ∠MRQ = 147° ...equation 4
Insert equation 4 in 3
3x° +3y° + 147° = 360°
3x +3y = 360 - 147
3x +3y = 213
3(x+y) = 3(71)
x+y = 71°
∠JQP = ∠RQK = 2y° (vertical angles are equal)
∠LRS = ∠QRK = 2x° (vertical angles are equal)
∠QRK + ∠RQK + ∠QKR = 180° (sum of angles in a triangle)
2x+2y + ∠QKR = 180
2(x+y) + ∠QKR = 180
2(71) + ∠QKR = 180
142 + ∠QKR = 180
∠QKR = 180 - 142
∠QKR = 38°
∠JKL = ∠QKR = 38°
Well for minimum wage it would be $7.25 but it depends on the state
X^2 + y^2 = 25
2x + y = -5.....y = -2x - 5
x^2 + (-2x - 5)^2 = 25
x^2 + (-2x - 5)(-2x - 5) = 25
x^2 + (4x^2 + 20x + 25) = 25
5x^2 + 20x + 25 = 25
5x^2 + 20x = 0
5x(x + 4) = 0
5x = 0 2(0) + y = -5
x = 0 y = -5
x + 4 = 0 2(-4) + y = -5
x = -4 -8 + y = -5
y = 3
solutions are : x = 0, y = -5.....and x = -4, y = 3
Answer:
should be 104
Step-by-step explanation:
30 + 30 + 10 + 10 + 12 + 12
Answer:

Step-by-step explanation:
The y-intercept (when x = 0) on the quadratic graph is 3, the upper quadratic graph does not intersect the x-axis, and the endpoint of the function has an open circle that stops at x = 1. The starting point extends to the boundaries of the graph
Therefore, the piecewise function shown in the quadratic graph is given as follows;
f(x) = x² + 3; x < 1
The straight line graph has a negative slope and starts at x = 1, with a closed circle, extending to higher values of <em>x</em> to the boundaries of the graph
Two points on the straight line graph are (1, 3) and (5, -5), therefore, the slope is (-5 - 3)/(5 - 1) = -2
The equation of the function is y - 3 = -2·(x - 1)
∴ y = f(x) -2·x + 2 + 3 = -2·x + 5
f(x) = -2·x + 5; x ≥ 1