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4vir4ik [10]
3 years ago
11

How many unique triangles can be made where one angle measures 98° and another angle is an obtuse angle?

Mathematics
2 answers:
Naddika [18.5K]3 years ago
8 0
The answer is none hope it help
Marta_Voda [28]3 years ago
7 0
The answer is none for the question

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Help please? 39 points
victus00 [196]
You cannot assume the angles add to 90, but you know since BD is an angle bisector that ABD is equal to DBC, or x-5=2x-6

x=1. this is the correct solution to the equation but gives negative angles when plugged in which isn't possible. there must be something wrong work the question
6 0
3 years ago
Select the statements that are true based on the following given information. D = {x | x is a whole number} E = {x | x is a perf
likoan [24]
<h3>2 Answers: Choice C and Choice D</h3>

==========================================================

Explanation:

Let's write the roster notation of each set

D = {0, 1, 2, 3, ...} the dots indicate the pattern goes on forever

E = {0, 1, 4, 9, 16, 25} = list of perfect squares smaller than 36

F = {20, 22, 24, 26, 28, 30} = even numbers between 20 and 30

----------------

If we intersect sets D and E, we're looking for what numbers are in both sets at the same time. Therefore, D ∩ E = {0, 1, 4, 9, 16, 25} which is the exact same as set E. This is because all of set E is inside of set D. We say that set E is a subset of set D. So, D ∩ E = E.

Choice A is very close to being true. The problem is that 25 is missing from the set {1,4,9,16}. So this is why choice A is false.

----------------

Now let's intersect sets D and F. The numbers they have in common are {20, 22, 24, 26, 28, 30} which is exactly what set F is. So set F is a subset of set D. We can write D ∩ F = F in much the same way we can say D ∩ E = E.

D ∩ F = {12,14,16,18} is not true. A number like 12 is not between 20 and 30, so it cannot be in set F.

Choice B is false so we cross it off the list.

----------------

Choice C is true and here's why

D ∪ (E ∩ F) is the same as saying D ∪ G where G is the set of intersecting E and F together. In other words, G = E ∩ F

We don't really need to even worry about sets E, F or G. All that matters here is set D.

When we write D ∪ (E ∩ F) or D ∪ G, we're saying "a number is in set D, or it is in set G". If it is in D, then it's a whole number. Otherwise, it's in a subset of whole numbers.

Overall, D ∪ G and D ∪ (E ∩ F) form the entire set of whole numbers.

------------------

Choice D is true.

E and F have nothing in common

E = {0, 1, 4, 9, 16, 25}

F = {20, 22, 24, 26, 28, 30}

So intersecting them leads to the empty set. This is the set with nothing inside it, not even 0.

------------------

Choice F is false

These two sets below

E = {0, 1, 4, 9, 16, 25}

F = {20, 22, 24, 26, 28, 30}

union together to get

H = {0,1,4,9,16,20,22,24,25,26,28,30}

just toss all of the numbers together into one big set

Notice how each of these numbers are whole numbers, so they are part of set D. This means set H is also a subset of set D.

When we intersect sets D and H, we end up with set H. We do not simply end up with the set with 25 only inside it.

6 0
3 years ago
Martin supera en 22 años a su hijo ernesto Expresa algebraicamente de 2 maneras distintas esta Relacion
scZoUnD [109]

you got the wrong brainly i guess

8 0
2 years ago
Please answer this question correctly and i'll mark you as brainlilest! Thank you.
raketka [301]

Answer:

\frac{ {7}^{4} ( - 2) {}^{4} {11}^{3}  6}{7 {}^{3}    {2}^{2} {6}^{2} 11 {}^{2}   }  =  \frac{7 \times  {2}^{2}  \times 11}{6}  =  \frac{308}{6}  =  \frac{154}{3}   \\  \\  \frac{ {7}^{0} {12}^{0}  {17}^{0}  }{ {57}^{0}  + ( {42}^{0}  \times ( {13}^{0} ))}  =  \frac{1}{1 + 1}  =  \frac{ 1}{2}

4 0
2 years ago
In a horse race, the odds in favor of the first horse winning in an 8-horse race are 2 to 5. The odds against the second horse w
Nataly_w [17]

The probability that the first horse wins is 2/7. The probability that the second horse wins is 3/10. Since the events that the first horse wins and the second horse wins are shared exclusive, the probability that either the first horse or the second horse will win is :

2/7 + 3/10= 41/70

Hope this is correct.

4 0
3 years ago
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