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postnew [5]
3 years ago
14

The sides of a right triangle consist of two legs and a hypotenuse. If the two legs of a right triangle each have a length of 26

centimeters and the hypotenuse has a length of 57 millimeters, what is the perimeter of the triangle? *
Mathematics
2 answers:
Ksju [112]3 years ago
7 0
<span>The sides of a right triangle consist of two legs and a hypotenuse. If the two legs of a right triangle each have a length of 26 centimeters and the hypotenuse has a length of 57 millimeters, what is the perimeter of the triangle.
</span>
Vikki [24]3 years ago
5 0
The sides of a right triangle consist of two legs and a hypotenuse. If the two legs of a right triangle each have a length of 26 centimeters and the hypotenuse has a length of 57 millimeters, what is the perimeter of the triangle? *A) 31.7 mmB) 83 mmC) 317 mm<span>D) 830 mm</span>
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You can only draw one unique isosceles triangle that contains an angle of 55°?
blsea [12.9K]

-- If the 55° angle is one of the two equal angles, then
the third angle is 70° .

-- If the 55° angle is the third angle, then each of the two
equal angles is 62.5° .

-- For either of these cases, there are an infinite number
of possible sets of side-lengths.

The statement in the question does not hold water.

6 0
3 years ago
Please help, thank you so much !!
Sliva [168]
Answer: x = 54

Explanation:

180 - 14 = 166

3x + 4 = 166
3x = 162
x = 162/3 = 54
6 0
2 years ago
In ΔABC, if m ∠A = m∠C, m∠B = ß (where ß is an acute angle), and BC = x, which expression gives the length of b, the side opposi
chubhunter [2.5K]

Answer:

The length of b is \sqrt{2x^2(1-cos\beta)}.

Step by step explanation:

Given information: In ΔABC, ∠A =∠C, ∠B = ß (where ß is an acute angle), and BC = x.

Since two angles are same therefore triangle ABC is an isosceles triangle and side AB and BC are congruent.

AB=BC=x

According to Law of cosine

b^2=a^2+c^2-2ac\cos B

b^2=x^2+x^2-2(x)(x)\cos \beta

b^2=2x^2-2x^2\cos \beta

b^2=2x^2(1-\cos \beta)

\sqrt{2x^2(1-cos\beta)}

Therefore the length of b is \sqrt{2x^2(1-cos\beta)}.

6 0
2 years ago
Find the geometric mean of four and 10
lawyer [7]

The Geometric mean of 4 and 10 is 6.32

<u>Explanation:</u>

Given:

Two numbers are 4 and 10

Geometric mean, GM = ?

We know,

GM = \sqrt[n]{a_1 X a_2}

Where,

n = 2

Substituting the value we get"

GM = \sqrt[2]{4 X 10} \\\\GM = \sqrt[2]{40} \\\\GM = 6.32

Thus, the Geometric mean of 4 and 10 is 6.32

4 0
3 years ago
What is the answer for this:
cricket20 [7]

Step-by-step explanation:

Okay! To do this you need to have x on one side (isolate the variable) and also combine like terms. Like terms are terms with the same exponent and variable, like 8x and 6x.

8x+1=-6x-2 To combine like terms here we have to move -6x to the other side of the equation by adding it (using inverse operations). 8x+6x=14x

14x+1=-2 now subtract 1 from both sides.

14x=-3

Divide both sides by 14 and x=-3/14

Brainliest? :)

7 0
3 years ago
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