Solve for x. (5x+17)+129+94+(7x-46)+(9x-11)= 540

Mathematics

1 answer:

marin [14]3 years ago

4 0

(5x+17)+129+94+(7x-46)+(9x-11)=540

5x+7x+9x+17+129+94-46-11=540

21x+146+94-57=540

21x+240-57=540

21x+183=540

-183 -183

21x=357

/21 /21

x=17

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hope it helps

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How many different whole number factors does 27 have?

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7 0

4 years ago

Would be very happy if you helped.Don’t spam guys

Reptile [31]

$\\ \sf\longmapsto 41sin\Theta=40$

$\\ \sf\longmapsto sin\Theta=\dfrac{40}{41}$

Now

$\boxed{\sf cos\Theta=\sqrt{1-sin^2\Theta}}$

$\\ \sf\longmapsto cos\Theta=\sqrt{1-\left(\dfrac{40}{41}\right)^2}$

$\\ \sf\longmapsto cos\Theta=\sqrt{1-\dfrac{1600}{1682}}$

$\\ \sf\longmapsto cos\Theta=\sqrt{\dfrac{1681-1600}{1681}}$

$\\ \sf\longmapsto cos\Theta=\sqrt{\dfrac{81}{1681}}$

$\\ \sf\longmapsto cos\Theta=\dfrac{9}{41}$

We know

$\boxed{\sf tan\Theta=\dfrac{Sin\theta}{Cos\Theta}}$

$\\ \sf\longmapsto \dfrac{tan\Theta}{1-tan^2\Theta}$

$\\ \sf\longmapsto \dfrac{\dfrac{sin\Theta}{cos\Theta}}{1-\dfrac{sin^2\Theta}{cos^2\Theta}}$

$\\ \sf\longmapsto \dfrac{\dfrac{\left(\dfrac{40}{41}\right)}{\left(\dfrac{9}{41}\right)}}{1-\dfrac{\left(\dfrac{40}{41}\right)^2}{\left(\dfrac{9}{41}\right)^2}}$