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Lelechka [254]
1 year ago
11

Pls answer for brainlest and 50 points PLEASE EXPLAIN In image

Mathematics
2 answers:
anzhelika [568]1 year ago
8 0
Answer: C) 163

Step-by-Step Solution:

In the Right Triangle formed to the extreme right, lets mark the angles as
∠1, ∠2 and ∠3.

Therefore, from the Figure :-

∠1 = 73°
∠2 = 90°

By Angle Sum Property :-

∠3 = 180 - (73 + 90)
∠3 = 180 - 163
=> ∠3 = 17°

The Angle which forms a Linear Pair with ∠3 is the Corresponding Angle of ∠r, and Corresponding Angles are Equal.

Therefore,
=> 180 - ∠3
= 180 - 17
=> 163°

Therefore, the Angle that forms the Linear Pair with ∠3 is 163°

This Angle is Corresponding to ∠r and hence they are Equal ie. ∠r = 163°

Hence, ∠r = 163°
AlexFokin [52]1 year ago
5 0

First, let's complete the angles in the triangle. Remember that the sum of the angles in a triangle is 180 degrees.

73 + 90 + x = 180

163 + x = 180

x = 17

So, the angle that completes the triangle is 17 degrees. If we look at that angle in the triangle and the one adjacent to it, we can see that those two angles form a linear pair (or are supplementary, both meaning that they add up to 180 degrees).

17 + x = 180

x = 163

So, 17's supplement is 163 degrees. The 163 degree angle corresponds with angle r, and corresponding angles are congruent.

Therefore, angle r is 163 degrees. The correct answer is option C.

Hope this helps!

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IRINA_888 [86]

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Answer:

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= [\frac{9}{2.6}  - \frac{2.5*2.5 }{2.5} ]^{2}

= [\frac{9}{2.6}  - \frac{2.5}{1} ]^{2}

*canceling 2.5 in numerator and denominator*

= [\frac{9-(2.5)(2.6)}{2.6} ]^2\\*Using L.C.M of 2.6 and 1 which comes out to be '2.6'= [\frac{9-(6.5)}{2.6} ]^2\\= [\frac{2.5}{2.6} ]^2\\*multiplying and dividing by '10'= [\frac{2.5*10}{2.6*10} ]^2\\= [\frac{25}{26} ]^2\\= \frac{25^2}{26^2}\\= \frac{625}{676}\\= 0.925

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The least or smallest common multiple of any two or more given natural numbers are termed as LCM. For example, LCM of 10, 15, and 20 is 60.

(b) [[\frac{3x^{a}y^{b}} {-3x^{a} y^{b} } ]^{3}    ] ^{2}

Answer:

[[\frac{3x^{a}y^{b}} {-3x^{a} y^{b} } ]^{3}] ^{2}\\

*using [x^{a}]^b = x^{ab}*

= [\frac{3x^{3a}y^{3b}} {-3x^{3a} y^{3b} }] ^{2}        

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= \frac{3x^{2*3a}y^{2*3b}} {-3x^{2*3a} y^{2*3b} }  \\= (-1)\frac{3x^{6a}y^{6b}} {3x^{6a} y^{6b} }\\[\tex]*taking -1 common, denominator and numerator are equal*[tex]= -(1)\frac{1}{1}\\= -1

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We can raise a power to a power

(x^2)4=(x⋅x)⋅(x⋅x)⋅(x⋅x)⋅(x⋅x)=x^8

This is called the power of a power property and says that to find a power of a power you just have to multiply the exponents.

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Step-by-step explanation:

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