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Tomtit [17]
3 years ago
10

A butlerfly has zero,one,more than onesymmetry lines​

Mathematics
1 answer:
raketka [301]3 years ago
7 0

Answer:

one line of symmetry

Step-by-step explanation:

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For the data in the table, does y vary directly with x? If it does, write an equation for the direct variation.
defon
I believe that it would be y = (13/8)x
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3 years ago
In rhombus ABCD, the diagonals AC and BDintersect at E. If AE=5 and BE=12, what is thelength of AB?
hram777 [196]

Diagonals that bisect each other are perpendicular and form a right angle.

AEB is a right triangle.

We can apply the Pythagorean theorem:

c^2 = a^2+b^2

Where c is the hypotenuse (longest side) and a and b the other 2 legs.

Replacing:

AB^2 = AE^2+BE^2

AB^2 = 5^2+12^2

AB^2 = 25+144

AB^2 = 169

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11 months ago
Amanda ate 1/3 of her mom's pie. she gave the remaining 2/3 to 4 friends, What fraction did each friend get.
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So she divided the 2/3 between 4 friends.

So each friend got 2/3 divided by 4, which gives:

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7 0
3 years ago
What are the solutions to the equation |x – 10| – 4 = 2x?
Yuliya22 [10]

|x – 10| – 4 = 2x

x - 10 - 4 = 2x

x - 14 = 2x

x - 2x = 14

-x = 14

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7 0
3 years ago
A short-wave radio antenna is supported by two guy wires, 150 ft and 180 ft long. Each wire is attached to the top of the antenn
Dafna1 [17]

The distance between the anchor points of the two guy wires holding radio antenna is 181 feet (to the nearest foot)

<h3>How to determine the distance between the two anchor points of the guy wire</h3>

The problem will be solved using SOH CAH TOA

let the distance between the 150 ft long guy wire and the radio antenna be x

let the distance between the 180 ft long guy wire and the radio antenna be y

cos 65° = x / 150

x = cos 65° * 150

x = 63.39 ft

The height of the antenna

sin 65° = height of antenna / 150

height of antenna = sin 65 * 150

height of antenna = 135.95

using Pythagoras theorem

(length of guy wire)² = (height of the antenna)² + (anchor distance)²

(anchor distance)² = 180² - 135.95²

anchor distance = √(180² - 135.95²)

anchor distance = 117.97

The anchor points distance apart

= 63.39 + 117.97

= 181 (to the nearest foot)

Learn more on Pythagoras theorem here:

brainly.com/question/29241066

#SPJ1

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1 year ago
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