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Dominik [7]
2 years ago
11

a construction worker needs to put a rectangular window in the side of a building. he knows from measuring that the top and bott

om of the window have a width of 5 feet and the sides have a length of 12 feet. he also measured one diagonal to be 13 feet. what is the length of the other diagonal?
Mathematics
2 answers:
Brilliant_brown [7]2 years ago
6 0

Width of the rectangular window = 5 feet

Length of the rectangular window = 12 feet.

Length of one of the diagonals is = 13 feet

As it is a rectangle, both the diagonals are equal, so other one will also be 13 feet. Or we can solve it by using Pythagoras theorem.

Let the supposed diagonal be = d

d^{2}=5^{2} +12^{2}

d^{2} =25+144

d^{2} =169

d=13

Hence, the length of the diagonal will be 13 feet.

Cloud [144]2 years ago
5 0
The other diagonal would be 13 feet as well because it is a rectangle.
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Take out a common factor between 3x and kx. That means use the distributive law to get what you normally would start with.

x(k + 3) = 4

Now divide by k + 3

x = 4/(k + 3)

That's as much as you can do with this question.



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1 year ago
A print shop makes bumper stickers for election campaigns. If x stickers are ordered (where x < 10,000), then the price per s
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Answer:

P(x) = (0.049x - 0.0000015x²)

Step-by-step explanation:

price per sticker is 0.14 − 0.000002x dollars

total cost of producing the order is 0.091x − 0.0000005x² dollars.

P(x) = profit = Revenue - Cost

Let the number of units of stickers made be x

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Cost of producing x units in the order = (0.091x − 0.0000005x²)

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P(x) = (0.049x - 0.0000015x²)

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3 0
2 years ago
Mr. Rico's small business is starting to make a profit. His costs last week were -$352. His profit last week was $22. How many t
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Answer:

The Answer is -16

Step-by-step explanation:

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5 0
2 years ago
In high-school 135 freshmen were interviewed.
timama [110]

Answer:

a) n(none) = 25

b) n(PE but not Bio) = 25

c) n(ENG but not both BIO and PE) = 55

d) n(students that did not take Eng or Bio) = 40

e) P( Students did not take exactly two subjects) = 0.65

Step-by-step explanation:

From the Venn diagram drawn:

a) Number of students that took none

n(Freshmen) = 135

n(all three) = 5

n (PE and Bio) = 10

n(PE and Eng) = 15

n(Bio and Eng) = 7

n (PE and Bio only) = 10 - 5 = 5

n(PE and Eng only) = 15 - 5 = 10

n(Bio and Eng only) = 7 - 5 = 2

n(PE only) = 35 - 5 - 5 - 10 = 15

n(Bio only) = 42 - 5 - 5 - 2 = 30

n(Eng only) = 60 - 10 - 5 -2 = 43

n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)

135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)

135 = 110 + n(none)

n(none) = 135 - 110

n(none) = 25

b)Number of students that too PE but not Bio

n(PE but not bio)= n(PE only) + n(PE and Eng only)

n(PE but not Bio) = 15 + 10

n(PE but not Bio) = 25

c) Number of students that took ENG but not both BIO and PE

n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10

n(ENG but not both BIO and PE) = 55

d) Number of students that did not take ENG or BIO

n( students that did not take Eng or Bio) = n(PE only) + n(none)

n(students that did not take Eng or Bio) = 15 + 25

n(students that did not take Eng or Bio) = 40

e) Probability that a randomly-chosen student from this group did not take exactly two subjects

n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)

n( Students that did not take exactly two subjects) = 15 + 30 + 43

n( Students that did not take exactly two subjects) = 88

P( Students did not take exactly two subjects) = 88/135

P( Students did not take exactly two subjects) = 0.65

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Answer:

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