explain why the sine of x is the sample regardless of which triangle is used to find it in the figure

1 answer:

4 0

Sine of x is used when to find the side of the triangle that does not have any right angle or is not a right triangle. This is according to the Law of Sines, which states that the sides of the triangle are to one another in the ratio as the sines of their opposite angles.

You might be interested in

E is the midpoint of AB. If AE = 4x +6 and EB = 10x - 12, what is AB?

zzz [600]

Answer:

Step-by-step explanation:

Since E is the midpoint, AE and EB are equal. We can write this as such,

4x+6=10x-12 if we solve from here we get 6x=18. Simplify to get x=3. sub in three for x in the AE equation, 4(3)+6 and solve. You'd get 18 as your answer. Double that to get the solution since AE= 1/2AB. 18*2= 36

AB=36

4 0

3 years ago

Determine the sequence

Shalnov [3]

Answer:

Add 3 each time. (for question #8)

Step-by-step explanation:

9+3=12

12+3=15

15+3=18

18+3=21

21+3=24

24+3=27

7 0

3 years ago

How do I do this and solve it

Nutka1998 [239]

Hey there! :D

Since the sides are equal in length, we know this is an isosceles triangle. It will have two equal angles.

We already know the vertex is 80 degrees, so the bottom two angles will be equal in measure. All angles in a triangle equal 180 degrees

180-80= 100

100/2= 50

x= 50

I hope this helps!
~kaikers

4 0

3 years ago

Read 2 more answers

Solve.

Ksenya-84 [330]

Solve.

x² + 4x + 4 = 18

A x=−4±3√ 2

B x=2±3√ 2

C x=4±9√ 2

D x=−2±3√2

 $x^2 + 4x + 4 = 18 \\ \\ x^2+4x+4-18= 0 \\ \\ x^2+4x-14=0 \\ \\ x_1_y_2= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} \qquad a= 1\qquad b= 4\qquad c= -14 \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{4^2-4(1)(-14)} }{2(1)} \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{16-(-56)} }{2} \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{72} }{2} \\ \\ \\ x_1= \dfrac{-4+ \sqrt{72} }{2} \qquad\qquad x_2= \dfrac{-4+ \sqrt{72} }{2}\\ \\ \\ x_1= \dfrac{-4+ \sqrt{2^3*3^2} }{2} \qquad\qquad x_2= \dfrac{-4- \sqrt{2^3*3^2} }{2}$

 $x_1= \dfrac{-4+2*3 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4- 2*3\sqrt{2} }{2} \\ \\ \\ x_1= \dfrac{-4+6 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4- 6\sqrt{2} }{2} \\ \\ \\ x_1= \dfrac{-4}{2} + \dfrac{6 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4}{2}- \dfrac{ 6\sqrt{2} }{2} \\ \\ \\ x_1= -2 + 3 \sqrt{2} \qquad\qquad\quad x_2= -2- 3\sqrt{2} \\ \\ \\ \boxed{x= -2\pm 3\sqrt{2} } \to D)$

7 0

3 years ago

Read 2 more answers

Change to fraction 336%

noname [10]

84/25 (84/25 = 3.36)
Hope to help
:D

3 0

3 years ago

Read 2 more answers