The value of cos A is √(1 + x²)/ (1 - x²) /√1 + x
<h3>Trigonometric ratios</h3>
It is important to note that
sin A = opposite/ hypotenuse
cos A = adjacent/ hypotenuse
Then,
opposite = 
Hypotenuse = 
Let's find the adjacent side using the Pythagorean theroem
cos A = x/hypotenuse
cos A = √(1 + x²)/ (1 - x²) /√1 + x
Thus, the value of cos A is √(1 + x²)/ (1 - x²) /√1 + x
Learn more about trigonometric identities here:
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Let's rephrase this:
Here:
There are 3 geometric transformations:
1. Rotation
Shapes are rotated or turned around an axis.
2. Reflection
Shapes are flipped across an imaginary line to make mirror images.
3. Translation
Shapes are slid across the plane.
Answer:
Step-by-step explanation:
First factor the second binomial
38^9 - 38^8 = 38^9 (38 - 1 ) = 38^9(37)
So there's your 37. This whole expression is divisible by 37
Now do the first binomial
36^5 - (36^4)*6
36^4(36 - 6)
36^4(30) and there's your thirty.
That first term is going to cause a bit of trouble showing that 36^4 * 6 = 6^9
6(36^4) = 6(6^2)^4 = 6^1 * 6^8 = 6^9
So this factors into 30*(36^4)*37*38^9
since a square must be the same length on all sides to be a correct square, you can legit just multiply the length of one side times four to get the area.
3.2 x 4 = 12.6
Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
