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2 years ago

Could anyone help with this, im terrible at geometry haha

Mathematics

1 answer:

Natalija [7]2 years ago

7 0

Answer:

m<ACD = 90

m<CDB = 126

m<EDB = 54

m<CDE = 180

Step-by-step explanation:

Well you have the right girl! *In honors geometry you can trust me!

Ok first thing...We already know

B = 54

A = 90

D = 54 (It's the Alternate interior angle of B)

C = 90

m<ACD = 90 (C = 90)

m<CDB = 126 (D = 54 ------> 180-54= 126)

m<EDB = 54 (Alternate Interior angle of B)

m<CDE = 180 (It is a straight line)

Hope this helps ya!!

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Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

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Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

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<h3>Answer: 24 feet (Choice D)</h3>

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Explanation:

Refer to the diagram below. The goal is to find x, which is the horizontal distance from the base of the tree to the swing set.

Focus on triangle BCD.

The angle B is roughly 30.26 degrees, and this is the angle of depression. This is the amount of degrees Emir must look down (when starting at the horizontal) to spot the swing set.

We know that he's 14 ft off the ground, which explains why AB = CD = 14.

The goal is to find BC = AD = x.

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Again, keep your focus on triangle BCD.

We'll use the tangent ratio to say

tan(angle) = opposite/adjacent

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tan(30.26) = 14/x

x*tan(30.26) = 14

x = 14/tan(30.26)

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Assuming the mass of the square is x, the equation for the first side is:

= (3 x mass of circle) + (2 x mass of triangle) + (6 x mass of square)

= ( 3 x 2) + ( 2 x 4) + ( 6 × x)

= 6 + 8 + 6x

Mass of other side:

= (2 x mass of circle) + (5 x mass of triangle) + (3 x mass of square)

= ( 2 x 2) + ( 5 x 4) + ( 3 × x)

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