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zaharov [31]
3 years ago
15

If triangle JKL is classified as isosceles, which statement is true?

Mathematics
1 answer:
VMariaS [17]3 years ago
6 0

Answer: At least two sides are congruent

Step-by-step explanation:

Here is the complete question:

If a triangle JKL is classified as isosceles, which statement is true?

a. At least two sides are congruent.

b. Two sides are perpendicular.

c. All three sides are congruent.

d. Two sides are parallel.

An isosceles triangle is a triangle that has at least two sides that are equal. An isosceles triangle has two equal sides and also two equal angles.

An angle is said to be congruent if it has the same angle either in degrees or radians. Such angles don't necessarily have to point in same direction. Therefore, it is true that at least two sides are congruent.

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If X²⁰¹³ + 1/X²⁰¹³ = 2, then find the value of X²⁰²² + 1/X²⁰²² = ?​
enyata [817]

Step-by-step explanation:

\bf➤ \underline{Given-} \\

\sf{x^{2013} + \frac{1}{x^{2013}} = 2}\\

\bf➤ \underline{To\: find-} \\

\sf {the\: value \: of \: x^{2022} + \frac{1}{x^{2022}}= ?}\\

\bf ➤\underline{Solution-} \\

<u>Let us assume that:</u>

\rm: \longmapsto u =  {x}^{2013}

<u>Therefore, the equation becomes:</u>

\rm: \longmapsto u +  \dfrac{1}{u}  = 2

\rm: \longmapsto \dfrac{  {u}^{2} + 1}{u}  = 2

\rm: \longmapsto{u}^{2} + 1 = 2u

\rm: \longmapsto{u}^{2} - 2u + 1 =0

\rm: \longmapsto  {(u - 1)}^{2} =0

\rm: \longmapsto u = 1

<u>Now substitute the value of u. We get:</u>

\rm: \longmapsto {x}^{2013}  = 1

\rm: \longmapsto x = 1

<u>Therefore:</u>

\rm: \longmapsto {x}^{2022}  +  \dfrac{1}{ {x}^{2022} }  = 1 + 1

\rm: \longmapsto {x}^{2022}  +  \dfrac{1}{ {x}^{2022} }  = 2

★ <u>Which is our required answer.</u>

\textsf{\large{\underline{More To Know}:}}

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)³ = a³ + 3ab(a + b) + b³

(a - b)³ = a³ - 3ab(a - b) - b³

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

(x + a)(x + b) = x² + (a + b)x + ab

(x + a)(x - b) = x² + (a - b)x - ab

(x - a)(x + b) = x² - (a - b)x - ab

(x - a)(x - b) = x² - (a + b)x + ab

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B80%7D%7B100%7D%20%20%3D%20" id="TexFormula1" title=" \frac{80}{100} = " alt=" \f
alukav5142 [94]
Answer: 0.8 or 80% (either way is correct)

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(Hope this helps!)
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Answer:

D

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in first step we must go on this way

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and D has this step

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