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LiRa [457]
3 years ago
6

1) On a city map drawn on a coordinate plane, the city park is located at (20, 4) and the school is at (20, 9). What is the dist

ance between the park and the school in the map units?
5 units

11 units

13 units

16 units

2) A rectangular plot of land is represented on a coordinate plane where each unit is one foot. The vertices are at (50, 20), (50, 90), (100, 20) and (100, 90). What is the perimeter of the rectangular plot?

120 feet

240 feet

520 feet

3500 feet
Mathematics
2 answers:
weqwewe [10]3 years ago
6 0
1) Since the x coordinates don't change, the distance is just delta y.

9-4=5 units

2)

w=dx=100-50=50

h=dy=90-20=70

p=2(w+h) so

p=2(70+50)

p=240 feet
Marina CMI [18]3 years ago
4 0
1) is 5 units which can be found by graphing
2) I got 240 by graphing this one as well
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Answer:

roots: 1 and 3

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Step-by-step explanation:

2 roots: p and p+2

(x-p) (x-p-2) = x² - 4x + k

x² -2px -2x + p² + 2p = x² - 2 (p+1)x + (p² + 2p) = x² -4x + k

-2 (p+1) = -4

p+1 = 2

p = 1   ... root 1

p' = 1+2 = 3   ... root 2

k = p² + 2p = 3

check: (x-1) (x-3) = x² - 4x + 3 = x² - 4x + k

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3 years ago
A ladder is leaning up against a house. The ladder is 20 feet long and the base of the ladder is 12 feet away from the house. Ho
MrRa [10]

Answer:

16 feet

Step-by-step explanation:

The length of the ladder=20 feet

Distance from the base of the ladder to the house = 12 feet

You will notice that a wall is vertical and the ladder makes an angle with the horizontal ground(making it the hypotenuse). This is a right triangle problem.

To find the how far up the house can the ladder can reach, we simply find the third side of the right triangle.

From Pythagoras theorem

Hyp^2=Opp^2+Adj^2\\20^2=12^2+Adj^2\\Adj^2=400-144\\Adj^2=256\\Adj=\sqrt{256}=16

The third side of the right triangle is 16. Therefore the ladder leans 16 feet from the ground.

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Suppose there is a pile of quarters dimes and pennies with a total value of $1.07 how much of each coin can be present without b
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since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0



4 0
3 years ago
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