Find the value of x and y.

Need some help with this question please

Serggg [28]

Answer:

$\frac{15}{17}$

Step-by-step explanation:

$\cos( \alpha ) = \frac{ad}{hip} \\ \\ \sin( \alpha ) = \frac{op}{hip}$

Pythagorean theorem

hip² = op² + ad²

17² = x² + 8²

x² = 17² - 8²

$x \: = \sqrt{{17}^{2} - {8}^{2} } \\ x = \sqrt{289 - 64} \\ x = \sqrt{225} \\ x = 15$

$\sin( \alpha ) = - \frac{15}{17}$

$\sin( - \alpha ) = \frac{15}{17}$

4 0

3 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
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7 0

3 years ago

Please I need help it’s an emergency

Fofino [41]

Answer:

the answer is 36

Step-by-step explanation:

beacuse if we use the pythagorean theorem for 9 and 12 we get 15 which means that the perimeter is P=9+15+12=36

8 0

3 years ago

Plz help its 7th grade distributive proprty

Marat540 [252]