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anygoal [31]
2 years ago
15

The image shows a quadrilateral inscribed in a circle. Select ALL true statements.

Mathematics
1 answer:
dimulka [17.4K]2 years ago
4 0

Answer:

c, d, e, g

Step-by-step explanation:

c. you already got it correct

d. since angle b is 100, arc AC must be two times that amount, which is 200

e. opposite angles of a quadrilateral are supplementary (adds up to 180). to find CDA, you have to do 180 minus 100 (which is the measure of angle B) to get 80

g. opposite angles of a quadrilateral are supplementary (adds up to 180). to find BCD, you need to do 180 minus 70 (which is the measure of angle A) to get 110

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$6.50 I guess, because I added $5 and $1.50 and got $6.50
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86% of what number is 77.4
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Read 2 more answers
#54 Simplify and write the answers using positive exponents only.
kramer
Gg easy

remember
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8 0
3 years ago
Solve for M <br> -3+m/ 9 = 10
sweet-ann [11.9K]

Step-by-step explanation:

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5 0
2 years ago
6. Two observers, 7220 feet apart, observe a balloonist flying overhead between them. Their measures of the
MaRussiya [10]

Answer:

The ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

Step-by-step explanation:

Let's call:

h the height of the ballonist above the ground,

a the distance between the two observers,

a_1 the horizontal distance between the first observer and the ballonist

a_2 the horizontal distance between the second observer and the ballonist

\alpha _1 and \alpha _2 the angles of elevation meassured by each observer

S the area of the triangle formed with the observers and the ballonist

So, the area of a triangle is the length of its base times its height.

S=a*h (equation 1)

but we can divide the triangle in two right triangles using the height line. So the total area will be equal to the addition of each individual area.

S=S_1+S_2 (equation 2)

S_1=a_1*h

But we can write S_1 in terms of \alpha _1, like this:

\tan(\alpha _1)=\frac{h}{a_1} \\a_1=\frac{h}{\tan(\alpha _1)} \\S_1=\frac{h^{2} }{\tan(\alpha _1)}

And for S_2 will be the same:

S_2=\frac{h^{2} }{\tan(\alpha _2)}

Replacing in the equation 2:

S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})

And replacing in the equation 1:

h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})=a*h\\h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}

So, we can replace all the known data in the last equation:

h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}\\h=\frac{7220 ft}{(\frac{1 }{\tan(35.6)}+\frac{1}{\tan(58.2)})}\\h=3579,91 ft

Then, the ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

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