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zubka84 [21]
3 years ago
9

There are three hallways at Wilson Middle School, as represented in the coordinate plane below. Four points are plotted on the c

oordinate plane to represent the end of each hallway.
Part B
The Wilson Middle School flagpole is 28 meters to the right of point D. Give the coordinates of the point that represent the location of the school's flagpole.

​

Mathematics
1 answer:
nordsb [41]3 years ago
7 0

Answer:

<h2>The coordinates are (50, 103)</h2>

Step-by-step explanation:

The three hallways are AB, BC, CD.

The co-ordinate of D is (50, 75).

The school's flagpole is 28 meters right of the point D.

Hence the y-coordinate for the flagpole will be same as D, that is 75.

The x-coordinate will be (28 + 75) = 103.

The location of the school's flagpole will be represented by (50, 103)

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Step-by-step explanation:

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2 years ago
Mary is the last person to get on a ferris wheel ride at a fair (the ride begins once she is on). The ferris wheel has a radius
Aleks [24]

Answer:

a) For sketch see attached picture.

 y=-8cos(\pi t)+10

b) Mary will be able to enjoy the scenery for a total of 1.68 min or about 101 seconds.

Step-by-step explanation:

a) In order to find the algebraic representation of the problem, we must start by drawing a sketch of what the problem looks like (see attached picture). Next, we can analyze it to see what function better fits the situation. In this case, since the ride starts at the lowest point, then we are talking about a negative cos function. A cos function has the followng general form:

y= Acos(\omega t - \phi) + C

where:

A= amplitude

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t=time

\phi= phase shift

C= vertical shift.

In this case the amplitude will be the radius of the ferris wheel, so:

A=-8

it's negative because it starts at the lowest point.

the angular speed is the angle the ride will move in a given amount of time. In this case:

\omega=\frac{\pi rad}{1 min}

\omega=\pi rad/min

There will be no phase shift for this problem.

the vertical shift is the height of the platform, so:

C= 2

so the algebraic expression is:

y=-8cos(\pi t)+10

b) In order to find the total time for which she will enjoy the ride we must start by building an inequality:

-8cos(\pi t)+10>12

and solve it for t. We can start by turning it into an equation, solve it for t and find the answers that can be found into its domain, in this case D=[0,4]

so we get:

-8cos(\pi t)+10=12

we subtract a 10 from both sides so we get:

-8cos(\pi t)=2

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cos(\pi t)=-\frac{1}{4}

Next, we take the inverse cosine to both sides and get different answers:

\pi t = 1.823+2\pi n

and

\pi t=4.46 + 2\pi n

we take both equations and divide them into pi so we get:

t = 0.58+2 n

and

t=1.42 + 2n

so now we find the possible answers that are between 0 and 4 minutes, this is for n=0 and 1, so we get the following times:

t={0.58, 1.42, 2.58, 3.42} min

we can now build our possible intervals for which the ferris wheel is higher than 12 m so we get the following intervals:

(0, 0.58) (0.58, 1.42) (1.42, 2.58) (2.58, 3.42) and (2.58, 4)

so we pick a test value for each of the intervals and test it on our equation. If the equation gives us a number that is greater than 12, then we have a valid interval. If we get a value that is less than 12, then we just discard that interval.

for (0, 0.58) we use 0.5 and get an answer of 10. We discard this interval.

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for (1.42, 2.58) we use a 2 and get an answer of 2. We discard this interval.

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and finally, for (2.58, 4) we use a 3.5 and get an answer of 10. We discard this interval

So we have two valid intervals now: (0.58, 1.42) and (1.42, 2.58). So now we can use them to find the time for which Mary will enjoy the scenery.

t_{1}=1.42min-0.58min=0.85min

t_{2}=2.58min-1.42min=0.85min

t_{total}=0.85min+0.85min

t_{total}=1.68min

so:

Mary will be able to enjoy the scenery for a total of 1.68 min or about 101 seconds.

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