Answer: 27.47
Step-by-step explanation:
(sin(70) * 10) / sin(20)) = 27.47
don't know much, but try that
Answer:
Statement 4 is incorrect.
Step-by-step explanation:
#4 is incorrect. This statement is assuming that both 3x+5 and 4x are equal, but the correct way to use them and achieve 180 (definition of supplementary) degrees is by writing it as: 3x+5 + 4x = 180. (I spaced the 3x+5 and 4x to make it clear that these are two angles added together.) With this equation, you will be able to find the x and the substitute it back in to complete the equation.


- <u>A </u><u>triangle </u><u>with </u><u>sides </u><u>11m</u><u>, </u><u> </u><u>13m </u><u>and </u><u>18m</u>

- <u>We</u><u> </u><u>have </u><u>to </u><u>check </u><u>it </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not</u><u>? </u>


According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.
<u>We </u><u>imply</u><u> </u><u>it </u><u>in </u><u>the </u><u>given </u><u>triangle </u><u>,</u>





<u>From </u><u>Above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>
The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

Answer:
The values for both of the missing sides is 8.
Step-by-step explanation:
The hypotenuse of a right triangle = a(square root of 2)
Therefore the other two missing sides equal a.
Therefore, the other two sides equal 8.
First, you need to rewrite the expression into binomial form, so you are working with two terms (as you world with a quadratic):
(x²)²-3(x²)-4=0
Now, you can place the x²s into brackets as the coefficient is now 1:
(x² )(x² )
Next, find out two numbers that add together to give you -3 and multiply to give -4 (these are the leftover integers after removing the x²s). These two numbers are -4 and 1.
Place the -4 and 1 into the brackets:
(x²-4)(x²+1)=0
Notice that the x²-4 is a difference of two squares, so can be further factorised into (x+2)(x-2)
This leaves you with a final factorisation of:
(x+2)(x-2)(x²+1)=0
Now we handle each bracket individually to obtain our four solutions for x:
x+2=0
x=-2
x-2=0
x=2
x²+1=0
x²=1
x=<span>±1</span>