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IrinaK [193]
3 years ago
13

Two friends are standing at opposite corners of a rectangular courtyard. The dimensions of the courtyard are 12 ft. by 25 ft. Ho

w far apart are the friends?
Mathematics
1 answer:
grigory [225]3 years ago
5 0

Answer:

37 feet

Step-by-step explanation:

Hope this helps

plz mark as brainliest!!!

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The only way I am can you I love my you can be do not know want r a to a party in for you every night before a federal court and then it will
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What are the four different ways to solve a quadratic equation? How can you determine the zeros, the vertex, and axis of symmetr
enyata [817]
Zeros: Where the parabola crosses the x-axis.

Vertex: The turning point of the parabola.

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3 years ago
Estimate the figure to the nearest whole number
olya-2409 [2.1K]
That means that, for example:

 if your number is 49 the nearest whole number would be 50.

so by saying estimate the figure to the nearest whole number , they are basically saying find your nearest whole number.

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3 years ago
An instructor gives a class a set of 10 problems and says that the nal exam will consist of a random selection of 5 of the probl
sweet-ann [11.9K]

Answer:

probability that he or she will answer at least 4 problems correctly = 0.5

Step-by-step explanation:

probability that he or she will answer at least 4 problems correctly is given as;

P (at least 4 correctly) = P (exactly 4 correctly) + P (all 5 correctly)

Since there are 10 problems. 4 is from the 7 problems chosen and 1 is from the 3 problems we can't figure out. Thus, P(exactly 4 correctly) = 7C4 x 3C1

Thus, P (at least 4 correctly) = [ (7C4) x (3C1) ] / (10C5)] + [(7C5)/(10C5)]

= [(35 x3)/252 ] + [21/252]

= [ 105/252 ] + [ 21/252 ]

= 126 / 252

= 0.5

8 0
3 years ago
If the maximum acceleration that is tolerable for passengers in a subway train is 1.74 m/s2 and subway stations are located 930
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Given that the subway stations are 930 m apart, the train have to be accerelated for half the distance and then decerelated for the rest of the distance.

Recall that the distance travelled by an object with an initial velocity, u, for a period of time, t, at an accereration, a, is given by
s=ut+ \frac{1}{2} at^2

But, we assume that the train accelerates from rest, thus
s=\frac{1}{2} at^2 \\  \\ \Rightarrow465=\frac{1}{2}(1.74)t^2 \\  \\ \Rightarrow t^2=534.48 \\  \\  \Rightarrow t=\sqrt{534.48}=23.12

The maximum speed is attained at half the center of the distance between subway stations (i.e. at distance = 465 m).

Thus, maximum speed = distance / time = 465 / 23.12 = 20.11 m/s.
6 0
3 years ago
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