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Radda [10]
3 years ago
7

Classify the triangle with the given side lengths as acute, right, obtuse, or not a triangle:

Mathematics
2 answers:
slavikrds [6]3 years ago
8 0

Answer:

It is not a right triangles because 15^2 + 13^2 is NOT EQUAL to 19^2

let x, y, and z be the angles inside the triangle.

Law of cosines says:

19^2 = 13^2 + 15^2 - 2(13)(15)cos x

361 = 169 + 225 - 390 cos x

-33 = -390 cos x

-33/-390 = cos x

11/130 = cos x

x = inverse cosine (11/130)

= 85.14609471338569.....

Per law of sines:

sin x/19 = sin y / 13

(13/19)sinx = sin y

inverse_sine ( (13/19) sin x) = y

y = 42.9810733896016548....

Per law of sines

sin x/19 = sin z / 15

z = inverse_sine( (15/19) sin x)

= 51.8728352950227805....

So it is an acute triangle and yes, these angles add up to 180

Over [174]3 years ago
7 0

Answer:

acute

Step-by-step explanation:

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If no (A) = 40%, no (B) = 30%, n (A∩B) = 20% then find n (<br> (A B)<br> using formula.
Vesnalui [34]

Answer:

solution:-We know that for any two finite sets A and B, n(A∪B)=n(A)+n(B)−n(A∩B).

Here, it is given that n(A)=20,n(B)=30 and n(A∪B)=40, therefore,

n(A∪B)=n(A)+n(B)−n(A∩B)

⇒40=20+30−n(A∩B)

⇒40=50−n(A∩B)

⇒n(A∩B)=50−40

⇒n(A∩B)=10

Hence, n(A∩B)=10

Step-by-step explanation:

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lina2011 [118]

\\ \sf\longmapsto \dfrac{1}{x-1}>4

  • Multiply x-1 on both sides

\\ \sf\longmapsto \dfrac{1}{x-1}\times (x-1)> 4(x-1)

\\ \sf\longmapsto 1>4(x-1)

\\ \sf\longmapsto 1> 4x-4

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\\ \sf\longmapsto 1+4> 4x

\\ \sf\longmapsto 5> 4x

  • Divide both sides by 4

\\ \sf\longmapsto \dfrac{5}{4}>x

\\ \sf\longmapsto x< \dfrac{5}{4}

Now

ATQ to the equation \bf \dfrac{1}{x-1} x should be greater than 1 .Because if it becomes 1 then the denominator will be 0 which is impossible.

Hence x>1

Thus

\\ \sf{:}\!\implies 1

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\\ \therefore\boxed{\bf x\epsilon \left(1,\dfrac{5}{4}\right)}

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