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son4ous [18]
3 years ago
12

Determine whether the set is well defined. The set of people who wear expensive perfume.

Mathematics
1 answer:
Tanya [424]3 years ago
3 0
Yes, the set <span>of people who wear expensive perfume is well-defined.</span><span> A set is said to be well-defined if its definition assigns it a unique value. Otherwise, the set is said to be not well-defined or ambiguous."W</span><span><span>ell-defined” means that any given object in the entire world, whether abstract or concrete is either an element of the set or isn't.</span> </span>
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A true-false test consists of 50 questions. how many does a student have to get right to convince you that he is not merely gues
umka2103 [35]
Hmmm. 20 or 30 so the teacher will count the rest,if a student guessed, wrong or right
7 0
3 years ago
The length of a rectangle is 6 yd more than twice the width x. The area is 416 yd2. Find the dimensions of the rectangle.
kati45 [8]

Answer:

13 yds * 32 yds

Step-by-step explanation:

A=area, l=length, w=width

A=w*l

l=6+2w

A=w*(6+2w)

A=6w+2w^2

416=6w+2w^2

416/2=(6w+2w^2)/2

208=3w+w^2

208-208=3w+w^2-208

w^2+3w-208=0

w^2+16w-13w-208=0

w(w+16)-13(w+16)=0

(w-13)(w+16)=0 ==> w+16=0 ==> w=-16, w can't be negative.

w-13=0

<u>w=13</u>

l=6+2w

l=6+2(13)

l=6+26

<u>l=32</u>

Dimensions: 13 * 32

8 0
1 year ago
Rational Numbers:Question 7
nevsk [136]

Answer:

<em>-5, -3 = Quadrant |||</em>

Step-by-step explanation:

-3, 5 = Quadrant ||

4, 2 = Quadrant |

2, -4 = Quadrant |||| (Quadrant |||| is also known as Quadrant IV)

<u><em>~ LadyBrain</em></u>

5 0
2 years ago
What is the domain of this function?
Amiraneli [1.4K]
Only the third one make sense I believe
7 0
3 years ago
In the right triangle shown, \angle A = 30^\circ∠A=30∘angle, A, equals, 30, degrees and AB = 8AB=8A, B, equals, 8.
Bogdan [553]

Answer:

BC=4 units

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

we know that

In the right triangle ABC

sin(A)=\frac{BC}{AB} ----> by SOH (opposite side divided by the hypotenuse)

we have

AB=8\ units\\A=30^o

substitute the given values

sin(30^o)=\frac{BC}{8}

solve for BC

BC=sin(30^o)(8)

Remember that

sin(30^o)=\frac{1}{2}

so

BC=(\frac{1}{2})(8)=4\ units

3 0
3 years ago
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