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Luden [163]
3 years ago
8

Angle r° = 2w°. What is the measure of angle r°?

Mathematics
Poor dude in need of advice
2 years ago
Answer doesn't work very well when instead of 127, it's 111
1 answer:
stira [4]3 years ago
8 0

Answer:

Option B. 286 degrees

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

Find the measure of angle BAD in triangle A

m\ BAD+127^o=180^o ----> by supplementary angles (form a linear pair)

m\ BAD=180^o-127^o=53^o

step 2

Find the measure of angle ABD in triangle A

we know that

The sum of the interior angles in any triangle must be equal to 180 degrees

so

m\ BAD+m\ ADB+m\ ABD=180^o

we have

m\ BAD=53^o

m\ ADB=90^o

substitute

53^o+90^o+m\ ABD=180^o

143^o+m\ ABD=180^o

m\ ABD=180^o-143^o

m\ ABD=37^o

step 3

Find the measure of angle w

we know that

m\ ABD+w=180^o ----> by supplementary angles (form a linear pair)

we have

m\ ABD=37^o

substitute

37^o+w=180^o

w=180^o-37^o

w=143^o

step 4

Find the measure of angle r

we have

r=2w ----> given problem

substitute the value of w

r=2(143^o)=286^o

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15 is the same as the product of 3 and a number. Which equation models this sentence?
adelina 88 [10]

Question

15 is the same as the product of 3 and a number. Which equation models this sentence?

15=3x

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15=3+x

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Answer:

<h2>15=3x</h2>

Step-by-step explanation:

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the answer is good


3 0
3 years ago
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Find the value of x then tell whether the side lengths form a Pythagorean triple 6 , 10 , x
masya89 [10]

Answer:

Explanation:

Use the Pythagorean theorem (

a

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400

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I've included a link that lists a couple of Pythagorean triples (a couple because they are infinitely many).

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3 0
2 years ago
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(a) If G is a finite group of even order, show that there must be an element a = e, such that a−1 = a (b) Give an example to sho
Dahasolnce [82]

Answer:

See proof below

Step-by-step explanation:

First, notice that if a≠e and a^-1=a, then a²=e (this is an equivalent way of formulating the problem).

a) Since G has even order, |G|=2n for some positive number n. Let e be the identity element of G. Then A=G\{e} is a set with 2n-1 elements.

Now reason inductively with A by "pairing elements with its inverses":

List A as A={a1,a2,a3,...,a_(2n-1)}. If a1²=e, then we have proved the theorem.

If not, then a1^(-1)≠a1, hence a1^(-1)=aj for some j>1 (it is impossible that a^(-1)=e, since e is the only element in G such that e^(-1)=e). Reorder the elements of A in such a way that a2=a^(-1), therefore a2^(-1)=a1.

Now consider the set A\{a1,a2}={a3,a4,...,a_(2n-1)}. If a3²=e, then we have proved the theorem.

If not, then a3^(-1)≠a1, hence we can reorder this set to get a3^(-1)=a4 (it is impossible that a^(-1)∈{e,a1,a2} because inverses are unique and e^(-1)=e, a1^(-1)=a2, a2^(-1)=a1 and a3∉{e,a1,a2}.

Again, consider A\{a1,a2,a3,a4}={a5,a6,...,a_(2n-1)} and repeat this reasoning. In the k-th step, either we proved the theorem, or obtained that a_(2k-1)^(-1)=a_(2k)

After n-1 steps, if the theorem has not been proven, we end up with the set A\{a1,a2,a3,a4,...,a_(2n-3), a_(2n-2)}={a_(2n-1)}. By process of elimination, we must have that a_(2n-1)^(-1)=a_(2n-1), since this last element was not chosen from any of the previous inverses. Additionally, a_(2n1)≠e by construction. Hence, in any case, the statement holds true.

b) Consider the group (Z3,+), the integers modulo 3 with addition modulo 3. (Z3={0,1,2}). Z3 has odd order, namely |Z3|=3.

Here, e=0. Note that 1²=1+1=2≠e, and 2²=2+2=4mod3=1≠e. Therefore the conclusion of part a) does not hold

7 0
3 years ago
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