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marissa [1.9K]
3 years ago
14

in mr kim's history class there are 16 male students and twelve female students. which ratio can be used to express female stude

nts to male students in simplest form
Mathematics
1 answer:
Doss [256]3 years ago
8 0

Anmale=16

female=12  

to find the ratio we can write it as

  16 : 12=4 : 3

for this we have to simplify both numbers with same table like in this we simplify both with table of 4

4*4=16

and

4*3=12

so ans =4:3


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Answer:

64√2 or 64 StartRoot 2 EndRoot

Step-by-step explanation:

A 45-45-90 traingle is a special traingle.  Let's say one of the leg of the triangle is x. The other one is also x because of the isosocles triangle theorem.  Therefore, using the pytagorean theorem, you find that x^2+x^2=c^2.  2(x)^2=c^2.  You then square root both sides and get c= x√2.  

Therefore, the two legs are x and the hypotenuse is x√2.  x√2=128 because the question says that the hypotenuse is 128.  Solve for x by dividing both sides by √2.  X=128/√2.  You rationalize it by multiplying the numberator and denominator of the fraction by √2.  √2*√2= 2.

X=(128√2)/2= 64√2 cm.

Since X is the leg, the answer would be 64√2

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Answer:

well that depends on how many pieces of gum are in one pack, once you have that info multiply it by 4.

Step-by-step explanation:

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So, we know the perimeter is 24 cm. and the top part is the left part, (x) + 2, but remember, we need to have it be a number that, multiplied by 2, is 24. So, using this formula, the length, (top / bottom) is 7, and x = 5. Because 5 + 5 = 10, and when length is +2, it adds up to 14. 10 + 14 = 24. So, length is 7, width, (left, right) is 5.
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The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?
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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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