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arsen [322]
3 years ago
6

The parking spaces in a city lot were created by painting rows of parallel line segments, then painting a line that cuts through

each of the parallel segments. A sketch of two parking spaces is shown below. 2 diagonal parallel lines are cut by a horizontal transversal. Where the first parallel line intersects the transversal, the top right angle is angle 1. Where the second parallel line intersects the transversal, the top left angle is angle 2. If m∠2 = 115°, what is m∠1?

Mathematics
2 answers:
Svetach [21]3 years ago
3 0

Answer:

65°

Step-by-step explanation:

The consecutive interior angles theorem states that if two parallel lines are cut by a transversal line then the resulting pair of consecutive angles formed are supplementary (that is they add up to 180°).

Given that the painting rows are parallel to each other and a painted line (transversal) cut the parallel lines forming consecutive interior angles m∠1 and m∠2. Therefore:

m∠1 + m∠2 = 180°

m∠1 + 115 = 180

m∠1 = 180 - 115

m∠1 = 65°

lapo4ka [179]3 years ago
3 0

Answer:

65

Step-by-step explanation:

just did it in edge  2020

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A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,
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Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of \pi/4.

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

d(\theta)=\frac{v_i^2sin(2\theta)}{g}

Where v_i is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is [0, \pi/2]. The critical  points of the function are those who make d'(\theta)=0:

d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}

d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...

The critical value inside the interval is \pi/4.

d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft

The second step is to find the values of the function at the endpoints of the interval:

d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft

The biggest value of f is gived by \pi/4, therefore \pi/4 is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of \pi/4.

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