+9488 Hahaha!! In my experience, "hectic" is a very fitting adjective to describe tar 
But there is another reason for my username. It is an anagram of what I wanted to be when I grew up.
+9488 a) We could use the Law of Cosines to find θ , but since △ABC is an isosceles triangle, we can split it into two
congruent right triangles by drawing a height from C to side AB, like this:
| sin( angle ) = opposite / hypotenuse | |
| sin( θ/2 ) = 12 / 20 |
|
| θ/2 = arcsin( 12/20 ) | |
| θ = 2arcsin( 12/20 ) |
|
| θ ≈ 1.287 |
b) arc length = (radius) * (angle measured in radians) ≈ (20 cm) * (1.287) ≈ 26 cm
c)
i) sector area = (angle measured in radians)/2 * (radius)2 ≈ (1.287)/2 * (20)2 cm2 ≈ 257.4 cm2
ii) shaded area = sector area - triangle area ...Do you know how to find the area of the triangle?
+9488 It appears there are four "zero-width space" characters in your LaTeX code. Two of them are between the 3/4 and the x which is why it doesn't look normal. I found that out by pasting your code into this website: https://www.soscisurvey.de/tools/view-chars.php
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As for your actual question, this is pretty much the method guest used:
34x+58=4x
And 34x=3x4 so we can rewrite it like this...
3x4+58=4x
To get a common denominator, we can multiply 3x4 by 22 which won't change its value.
22⋅3x4+58=4x
And 22⋅3x4=6x8 ←That's how it went from 3x to 6x
6x8+58=4x
Now that the fractions have a common denominator we can combine them.
6x+58=4x
Multiply both sides of the equation by 8
6x+5=32x
Subtract 6x from both sides.
5=32x−6x
5=26x
Divide both sides by 26
526=x
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Also, we could multiply both sides of the equation by 8 to begin with, like this:
34x+58 = 4x 8(34x+58) = 8(4x) 8⋅34⋅x + 8⋅58 = 8⋅4⋅x 6x + 5 = 32x 5 = 26x 526 = x_
