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3 years ago

The sum of mZAOB and mZBOC IS 128º. Find the measure of ZBOC.

Mathematics

2 answers:

amm18123 years ago

4 0

128 - 33 = 95
Answer C. 95

Arada [10]3 years ago

3 0

Answer:

Step-by-step explanation:

m∠AOB + m∠BOC = m∠AOC

m∠AOB is an acute angle. Because it's 33°, less than 90°

m∠AOC is 128° , so it's an obsute angle . Because it's more than 90°, less than 180°

To find m∠BOC , just subtract m∠AOB from m∠AOC

128° = 33° + m∠BOC

m∠BOC = 128° - 33°

m∠BOC = 95°

Option C

So, m∠BOC is an obsute angle . Because it's more than 90°, less than 180°

more about angles: Right angle = exactly 90° , Straight angle = exactly 180°,

Full angle = exactly 360°

Hope this helps ^-^

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3: distribute the negative. X + x + 1 = 7
4: combine like terms. 2x + 1 = 7
5: solve for x. Subtract 1 on both sides. 2x = 6
6: divide by 2 to get x by itself. X = 3
7: plug the new value of x into one of the ORIGINAL equations. 3 + y = -1
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Step-by-step explanation:

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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:

professor190 [17]

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Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is $\{1,2,3,4\}$ .

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Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

1R4 and 4R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
4R1 and 1R2, then 4 must be related with 2.

Therefore $\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\}$ is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

since 1R2, then 2 must be related with 1.
since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

1R2 and 2R1, then 1 must be related with itself.
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2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

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For be reflexive

1 must be related with 1,

2 must be related with 2,

3 must be related with 3,

4 must be related with 4

For be symmetric

since 1R2, 2 must be related with 1,
since 1R4, 4 must be related with 1.

For be transitive

Since 4R1 and 1R2, 4 must be related with 2,
since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

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Answer:

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Step-by-step explanation:

The options of the question are

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we have

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The vertex represent the minimum value

The quadratic equation in vertex form is

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a is a coefficient

(h,k) is the vertex

Convert the quadratic equation in vertex form

Factor 2 leading coefficient

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Complete the squares

$2(x^{2} -2x+1)-2-2=0$

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Rewrite as perfect squares

$2(x-1)^{2}-4=0$

The vertex is the point (1,-4)

Move the constant to the right side

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Step-by-step explanation:

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