The answer is (x+2)(6x-5) since 6x * x is 6x^2, (-5 * 2) is -10 and 12x-5x is 7x
Given that
z₁ = 15 (cos(90°) + i sin(90°))
z₂ = 3 (cos(10°) + i sin(80°))
we get the quotient z₁/z₂ by dividing the moduli and subtracting the arguments:
z₁/z₂ = 15/3 (cos(90° - 10°) + i sin(90° - 10°))
z₁/z₂ = 5 (cos(80°) + i sin(80°))
so that z₁ is scaled by a factor of 1/3 and is rotated 10° clockwise.
The product of two rational numbers is always rational because (ac/bd) is the ratio of two integers, making it a rational number.
We need to prove that the product of two rational numbers is always rational. A rational number is a number that can be stated as the quotient or fraction of two integers : a numerator and a non-zero denominator.
Let us consider two rational numbers, a/b and c/d. The variables "a", "b", "c", and "d" all represent integers. The denominators "b" and "d" are non-zero. Let the product of these two rational numbers be represented by "P".
P = (a/b)×(c/d)
P = (a×c)/(b×d)
The numerator is again an integer. The denominator is also a non-zero integer. Hence, the product is a rational number.
learn more about of rational numbers here
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Answer:
Circumscribed circle: Around 80.95
Inscribed circle: Around 3.298
Step-by-step explanation:
Since C is a right angle, when the circle is circumscribed it will be an inscribed angle with a corresponding arc length of 2*90=180 degrees. This means that AB is the diameter of the circle. Since the cosine of an angle in a right triangle is equivalent to the length of the adjacent side divided by the length of the hypotenuse:
To find the area of the circumscribed circle:
To find the area of the inscribed circle, you need the length of AC, which you can find with the Pythagorean Theorem:
The area of the triangle is:
The semiperimeter of the triangle is:
The radius of the circle is therefore 
The area of the inscribed circle then is 
.
Hope this helps!
