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kondaur [170]
1 year ago
12

The triangle above is isosceles and b> a. Which of the following must be FALSE?

Mathematics
1 answer:
kow [346]1 year ago
5 0

If the triangle is isosceles and b is greater than a then,

(c) AC = BC is false.

An isosceles triangle in geometry is one with at least two equal-length sides. It can be defined as having exactly two equal-length sides or as having at least two equal-length sides, with the equilateral triangle being an exception to the second definition.

The triangle is an isosceles triangle and angle b is greater than angle b.

For option (A),

AB = BC, therefore a = c which is possible.

For option (B),

AB = AC, therefore b = c which is also possible.

For option (C),

AC = BC, therefore a =b.

But we have b > a. Hence AB  = BC is false.

For option (D),

a = c, therefore AB = BC which is possible according to the properties of an isosceles triangle.

Option (C) is false.

Learn more about isosceles triangle here:

brainly.com/question/16294004

#SPJ9

The complete question is mentioned below:

Figure not drawn to scale. The triangle above is isosceles and b > a. Which of the following must be FALSE?

A) AB = BC

B) AB = AC

C) AC = BC

D) a = c

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Given:

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Find the standard deviation of the probability distribution.

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Step 2:

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Standard\text{ Deviation}=\sqrt{\sum_{i\mathop{=}0}^3(x_i-\mu)^2P_(x_i)}=(0-2)^2(0.25)+(1-2)^2(0.05)+(2-2)^2(0.15)+(3-2)^2(0.55)=4(0.25)+1(0.05)+0(0.15)+1(0.55)=1+0.05+0+0.55=1.60

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Answer:

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\tan \: 30 \degree =  \frac{x}{10 \sqrt{3} }  \\  \\  \frac{1}{ \sqrt{3} } =  \frac{x}{10 \sqrt{3} }  \\  \\ \sqrt{3} x = 10 \sqrt{3}  \\  \\ x =  \frac{10 \sqrt{3} }{ \sqrt{3} }  \\  \\ x = 10 \\  \\  \cos 30 \degree =  \frac{10 \sqrt{3} }{y}  \\  \\  \frac{ \sqrt{3} }{2}  =  \frac{10 \sqrt{3} }{y}  \\  \\ y =  10 \sqrt{3}  \times  \frac{2}{ \sqrt{3} }  \\  \\ y = 10 \times 2 \\  \\ y = 20

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