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trapecia [35]
3 years ago
14

the hypotenuse and a side of a right angled triangle are 13cm and5cm respectively. fined the length of the third side​

Mathematics
2 answers:
alexgriva [62]3 years ago
8 0

Answer:

<h2>Lenght of the third side =12</h2>

Step-by-step explanation:

hypotenuse = 13 \\ opposite = 5 \\ adjacent = x \\  {hyp}^{2}  =  {opp}^{2}  +  {adj}^{2}  \\

{13}^{2}  =  {5}^{2}  +  {x}^{2}  \\ 169 = 25 +  {x}^{2}  \\ 169 - 25 =  {x}^{2}  \\ 144 =  {x}^{2}

\sqrt{144}  =  \sqrt{ {x}^{2} }  \\ 12 = x

eduard3 years ago
5 0

Answer:

If the third side is the hypotenuse, then (using the Pythagorean Theorem) the square of its length is the sum of the squares of the other two sides:  132+52 , so the length of the hypotenuse is √194=13.92838827718412...  

However, it is also possible to have a right triangle with a hypotenuse of 13 cm and one side of 5 cm, in which case the third side would be:  √132−542=12.  

Of course, the second answer may seem “nicer” but both are correct and the question is ambiguous.

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BabaBlast [244]

Answer:

<h3>           sum of the roots:    x_1+x_2=4\frac12</h3><h3>           product of the roots:     x_1x_2=-5             </h3><h3> Step-by-step explanation:</h3>

2x^2-9x-10=0\quad\implies\quad a=2\,,\ b=-9\,,\ c=-10

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From Vieta's formulas applied to quadratic polynomial we have:

if    b^2-4ac\geqslant0   then

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3 years ago
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Answer:

(a) The probability that a randomly selected U.S. adult uses social media is 0.35.

(b) The probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c) The probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = an US adult who does not uses social media.

<em>Y</em> = an US adult between the ages 18 and 29.

<em>Z</em> = an US adult between the ages 30 and above.

The information provided is:

P (X) = 0.35

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P (Y ∪ X') = 0.672

(a)

Compute the probability that a randomly selected U.S. adult uses social media as follows:

P (US adult uses social media (<em>X'</em><em>)</em>) = 1 - P (US adult so not use social media)

                                                   =1-P(X)\\=1-0.35\\=0.65

Thus, the probability that a randomly selected U.S. adult uses social media is 0.35.

(b)

Compute the probability that a randomly selected U.S. adult is aged 18–29 as follows:

P (Adults between 18 - 29 (<em>Y</em>)) = 1 - P (Adults 30 or above)

                                            =1-P(Z)\\=1-0.78\\=0.22

Thus, the probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c)

Compute the probability that a randomly selected U.S. adult is 18–29 and a user of social media as follows:

P (Y ∩ X') = P (Y) + P (X') - P (Y ∪ X')

                =0.22+0.65-0.672\\=0.198

Thus, the probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.

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