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Harman [31]
2 years ago
11

Which biconditional is NOT a good definition?

Mathematics
2 answers:
Lady_Fox [76]2 years ago
8 0
The answer is D because and obtuse angle is 91 degrees or more
Vladimir79 [104]2 years ago
6 0
<h2>Answer:</h2>

Option: d is the correct answer.

  d.   An angle is obtuse if and only if its measure is greater than 180°      

<h2>Step-by-step explanation:</h2>

<u>Acute angle--</u>

An angle is acute if and only if the measure of the angle is strictly less than 90 degree.

<u>Right angle--</u>

An angle is right angle if and only if the measure of the angle is exactly equal to 90 degree.

<u>Obtuse angle--</u>

An angle is said to be obtuse if and only if the measure of the angle is strictly greater than 90 degree and strictly less than 180 degree.

<u>Congruent angles--</u>

Two angles are said to be congruent if and only if both the angles have the same measure.

Hence, the biconditional which is NOT a good definition is:

                                    Option: d

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Hi, this question is a little unclear.  If you want to know what percentage he saved, it would be 30%, and to convert that percentage into a decimal would be a decimal of 0.30

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Evaluate.
aleksandrvk [35]

Answer: -2 4/15

Step-by-step explanation: To properly subtract, we need to find a common denominator. We can list the multiples of 3 and 5 to find the common denominator:

3: 3, 6, 9, 12, 15

5: 5, 10, 15


The first common multiple we see is 15.

Calculation:

3 · 3 = 9

9 + 1 = 10

3 1/3 = 10/3

5 x 5 = 25

25 + 3 = 28

-5 3/5 = -28/5

Now to convert into fifteen as the denominator:

-28/5 = -28 x 3/5 x 3 = -84/15

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Now to subtract accordingly:

50/15 - 84/15 = -34/15

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What is the domain of the square root function graphed below?
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Answer:

The domain of the graph must be x\ge \:-4.

Therefore,

x\ge \:-4\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\ge \:-4\:\\ \:\mathrm{Interval\:Notation:}&\:[-4,\:\infty \:)\end{bmatrix}

Hence, option a is true.

Step-by-step explanation:

From the graph, it is clear that the graph is heading towards positive infinity from x=-4.

The point x=-4 is included in the graph as the starting point of the graph i.e. x=4 is showing a closed circle on x=4, and heading towards positive infinity onward.

i.e. [-4, ∞)

Hence, the domain of the graph must be x\ge \:-4.

Therefore,

x\ge \:-4\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\ge \:-4\:\\ \:\mathrm{Interval\:Notation:}&\:[-4,\:\infty \:)\end{bmatrix}

Hence, option a is true.

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