The solution to the problem is as follows:
<span>4 sqrt(16x^2 y^7)
Protip: sqrt(abc) = sqrt(a)sqrt(b)sqrt(c). Therefore,
4sqrt(16x^2y^7) = 4sqrt(16)sqrt(x^2)sqrt(y^7)
We can deal with sqrt(16); it is equal to 4.
4(4) sqrt(x^2)sqrt(y^7)
16sqrt(x^2)sqrt(y^7)
Next, it is a known fact that sqrt(z^2) is the same as |z|, or the absolute value of z. That means
sqrt(x^2) = |x|
16|x| sqrt(y^7)
Split y^7 as y^6 * y
16|x| sqrt(y^6 * y)
16|x| sqrt(y^6) sqrt(y)
16|x| sqrt([y^3]^2) sqrt(y)
16|x| |y^3| sqrt(y)
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Each angle in the first triangle is 60 degrees because all sides are equal. the missing angles in the second triangle are 40 degrees each because the triangle is an isosceles triangle in which two angles are equivalent
If we draw a rhombus OABC on a coordinate grid with one vertex at the point O (0,0)
and let its height be 1 then (using the Pythagoras theorem) the coordinates of the 4 points will be as follows:-
O = (0,0) , A = (1, 1), B = (sqrt2+ 1 , 1) , C = (sqrt2, 0)
Slope of the diagonal OB = 1 / (sqrt2 + 1)
Slope of the other diagonal AC = - 1 / (sqrt2 - 1)
Product of the slopes = - 1 / ( sqrt2 + 1)(sqrt2 - 1) = -1 / 1 = -1
This proves that the diagonals are perpendicular.
Answer:
x = 55°
Step-by-step explanation:
x = 180° - 85° - 40° = 55°
12/3 = x/8
Cross multiply
3x = 96
Eliminate
3x/3 = 96/3
x = 32
