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lakkis [162]
2 years ago
14

Could anyone help me? I will mark you as brain liest who ever helps me!​

Mathematics
2 answers:
PolarNik [594]2 years ago
6 0

Answer:

1)163.28 (3.14*52)

2)13ft 6in (163.28/12)

3)129 times

4) 31 times

Step-by-step explanation:

1) multiply 52 times pie(3.14)

2)divide 163.28(your total) by 12

3) 1760 divided by 52

4) 1760 divided by 56.52

Westkost [7]2 years ago
3 0

this is for the other 2 ones

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Step-by-step explanation:

So let's think about what values you can take square root of...

anything but negative values!

So the first one is not correct because you can also take square root of 0,

squareroot of 0 is 0.

So because of this, C and D are both correct, E is correct because root  x is an increasing function. B is incorrect because (0,0) is an intercept.

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Scientists researching amphibians put tags on 30 frogs in a wetland area. Later, the scientists captured 520 frogs, of which 12
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3 years ago
One of the vertices of an equilateral triangle is on the vertex of a square and two other vertices are on the not adjacent sides
Elina [12.6K]
<h2>Answer:</h2>

<em> The side of the triangle is either 38.63ft or 10.35ft</em>

<h2>Step-by-step explanation:</h2>

This problem can be translated as an image as shown in the Figure below. We know that:

  • The side of the square is 10 ft.
  • One of the vertices of an equilateral triangle is on the vertex of a square.
  • Two other vertices are on the not adjacent sides of the same square.

Let's call:

Since the given triangle is equilateral, each side measures the same length. So:

x: The side of the equilateral triangle (Triangle 1)

y: A side of another triangle called Triangle 2.

That length is the hypotenuse of other triangle called Triangle 2. Therefore, by Pythagorean theorem:

\mathbf{(1)} \ x^2=100+y^2

We have another triangle, called Triangle 3, and given that the side of the square is 10ft, then it is true that:

y+(10-y)=10

Therefore, for Triangle 3, we have that by Pythagorean theorem:

(10-y)^2+(10-y)^2=x^2 \\ \\ 2(10-y)^2=x^2 \\ \\ \\ \mathbf{(2)} \ x^2=2(10-y)^2

Matching equations (1) and (2):

2(10-y)^2=100+y^2 \\ \\ 2(100-20y+y^2)=100+y^2 \\ \\ 200-40y+2y^2=100+y^2 \\ \\ (2y^2-y^2)-40y+(200-100)=0 \\ \\ y^2-40y+100=0

Using quadratic formula:

y_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ y_{1,2}=\frac{-(-40) \pm \sqrt{(-40)^2-4(1)(100)}}{2(1)} \\ \\ \\ y_{1}=37.32 \\ \\ y_{2}=2.68

Finding x from (1):

x^2=100+y^2 \\ \\ x_{1}=\sqrt{100+37.32^2} \\ \\ x_{1}=38.63ft \\ \\ \\ x_{2}=\sqrt{100+2.68^2} \\ \\ x_{2}=10.35ft

<em>Finally, the side of the triangle is either 38.63ft or 10.35ft</em>

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