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Travka [436]
3 years ago
8

The science fair judges will be science and volunteers each judge will only have time to view 5 projects there are 133 science t

eachers what is the fewest number of volunteers needed to have enough judges for all of the projects
Mathematics
1 answer:
antiseptic1488 [7]3 years ago
8 0
5 times amount of people has to be bigger than 2112

5(133+v)≥2112
distribute
665+5v≥2112
minus 665 from both sides
5v≥1447
divide both sides by 5
v≥289.4
yo can't have 0.4 volunterer
v≥290

has to have at least 290 volunteers
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What is greater, 85/100 or 21/25?
Elodia [21]


85/100 is greater

85/100=.85 or 85%

21/25=.84 or 84%

.85 is greater than .84 so 85/100 > 21/25

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3 years ago
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What is the percent increase in 300 and 375
inysia [295]

Answer:

25

Step-by-step explanation:

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2 years ago
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Can someone help me to find length CD
uysha [10]

Answer:

CD = 3.602019190339

Step-by-step explanation:

CD = DA - CA

DA = DB×Cos(29) = 18.7×cos(29) = 16.355388523507

BA = BA×cos(43) = 18.7×cos(43) = 13.676314220278

CA = BA÷tan(47) = 13.676314220278÷tan(47) = 12.753369333168

Then

CD = 16.355388523507 - 12.753369333168 = 3.602019190339

7 0
3 years ago
The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?
Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

A(-6, 2)) = (x_1, y_1)

B(2, 6) = (x_2, y_2)

AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}

AB = \sqrt{(8)^2 + (4)^2}

AB = \sqrt{64 + 16} = \sqrt{80}

AB = 8.9 (nearest tenth)

Distance between B(2, 6) and C(7, 1):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

B(2, 6) = (x_1, y_1)

C(7, 1) = (x_2, y_2)

BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}

BC = \sqrt{(5)^2 + (-5)^2}

BC = \sqrt{25 + 25} = \sqrt{50}

BC = 7.1 (nearest tenth)

Distance between C(7, 1) and D(3, -5):

CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

C(7, 1) = (x_1, y_1)

D(3, -5) = (x_2, y_2)

CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}

CD = \sqrt{(-4)^2 + (-6)^2}

CD = \sqrt{16 + 36} = \sqrt{52}

CD = 7.2 (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

D(3, -5) = (x_1, y_1)

A(-6, 2) = (x_2, y_2)

DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}

DA = \sqrt{(-9)^2 + (7)^2}

DA = \sqrt{81 + 49} = \sqrt{130}

DA = 11.4 (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

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3 years ago
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