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Allisa [31]
2 years ago
13

Suppose a triangle has sides a,b, and c with side c the longest side, and that a^2+b^2>c^2. Let theta be the measure of the a

ngle opposite the side of length c. Which of the following must be true. CHECK ALL THAT APPLY
A the triangle is not a right triangle
B theta is an acute angle
C The triangle in question is a right triangle
D costheta<0
Mathematics
1 answer:
andreyandreev [35.5K]2 years ago
6 0

Answer :

<u>A the triangle is not a right </u><u>triangle</u>

<u>D </u><u>costheta</u><u> </u><u><0</u>

The law of cosines says:

a²=b²+c²−2bccos(θ)

Rewriting:

2bccos(θ)=b²+c²−a²<0

So cos(θ)<0cos(θ)<0 since 2bc>02bc>0. Since 0<θ<π0<θ<π in any triangle,

π/2<θ<π

So:

1. θ is not an acute angle.

2 The triangle is not a right triangle. In a right triangle, one of the angles is 90 degrees and the other two are then less than 90 degrees. This triangle has an angle greater than 90 degrees.

3. cos(θ)<0 is true.

4. cos(θ)>0 is false -3 says cos(θ) is negative; a number can't be both positive and negative.

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A circle has a diameter of 10 centimeters. of a sector with a central angle of 50 degrees is cut out, what is the area of this s
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Answer:

D. 10.91 square centimeters

Step-by-step explanation:

Area of a sector of a circle = \frac{\theta}{360}\\ \pir^{2}

where \theta = 50^{o} C

radius (r) = half of diameter = 10/2 = 5 cm

Area  = \frac{50}{360} * \frac{22}{7} * 5^{2}

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Suppose you have a group of 10 children consisting of 4 girls and 6 boys. how many four -person teams can be chosen that consist
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4 years ago
Please help me ASAP which figure shows the conjecture is false??
Tema [17]

The easiest place to start is with the answer. The 3 by 4 rectangle (upper right) is the answer. P = 2(L + W) = 2(4 + 3) = 2*7 is 14.


14 is less than 16.


Now lets find out why the others can't be true.


The upper left has an area of 12 (6*2). So the area is find.

P = 2 * (L + W) = 2* (6 + 2) = 16 This would support the conjecture so it is not the answer you want.


The lower left has an area of 12 and a perimeter of 2*(12 + 1) = 26. That supports the conjecture. So you don't want that one either.


The lower right does not have an area of 12. It cannot help us at all.

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3 years ago
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