What is the measurements the for the angles

Mathematics

1 answer:

koban [17]3 years ago

5 0

The angle TQS is 74° and the angle SQR is 106°

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Check down the image that’s give below

gregori [183]

Solution:

$\frac{3x}{ {x}^{2} + 6x + 9} + \frac{x + 3}{ {x}^{2} - 9 }$

In the first fraction, we have to factorise the denominator using (a + b)² = a² + 2ab + b². And in the second fraction, we have to factorise the denominator using a² - b² = (a - b)(a + b) identity.

$= \frac{3x}{ {(x)}^{2} + 2(x)(3) + ( {3)}^{2} } + \frac{x + 3}{(x)^{2} - (3)^{2} } \\ = \frac{3x}{(x + 3) ^{2} } + \frac{x + 3}{(x + 3)(x - 3)}$

From the second fraction, cancel out from both sides (x + 3), the we get:

$= \frac{3x}{(x + 3) ^{2} } + \frac{1}{(x - 3)} \\ = \frac{3x(x - 3) + 1(x + 3) ^{2} }{(x + 3)^{2} (x - 3)} \\ = \frac{3x ^{2} - 9 x+ {x}^{2} + 6x + 9}{ {(x + 3)}^{2} (x - 3)} \\ = \frac{ {4x}^{2} - 3x + 9}{(x + 3)(x + 3)(x - 3)}$

Answer:

$\frac{ {4x}^{2} - 3x + 9}{(x + 3)(x + 3)(x - 3)}$

Hope you could understand.

If you have any query, feel free to ask.

4 0

2 years ago

15 points!!! GEOMETRY help please!! Questions attached!

marishachu [46]

9.
$\text{If QR is tangent to circle then m}\angle PQR=90^o$

$\text{We can use the converse of Pythagorean theorem}$

$8\ and\ 4\sqrt3\\\\check:\ 4^2+(4\sqrt3)^2=8^2\\16+16\cdot3=64\\16+48=64\\64=64\ -O.K.\\\\2\ and\ 4\sqrt2\\\\check:\ 4^2+2^2=(4\sqrt2)^2\\16+4=16\cdot2\\20=32\ -FALSE\\\\3\ and\ 5\\\\check:\ 3^2+4^2=5^2\\9+16=25\\25=25\ -O.K.\\\\2\ and\ 2\sqrt5\\\\check:\ 4^2+2^2=(2\sqrt5)^2\\16+4=4\cdot5\\20=20\ -O.K.\\\\6\ and\ 10\\check:\ 4^2+6^2=10^2\\16+36=100\\52=100\ -FALSE$

$\text{Answer:}\boxed{8\ and\ 4\sqrt3;\ 3\ and\ 5;\ 2\ and\ 2\sqrt5}$

10.
$\text{Look at the pictures}$

$\text{Therefore}\\\\C=2\cdot8+2\cdot16+2\cdot9+2\cdot6=16+32+18+12=78cm$

$Other\ method:\\C=2\cdot(8+16+6+9)=2\cdot39=78cm$