For each question we must apply the Pythagorean theorem and clear the value of x.
question 5
x = root (14 ^ 2 + 16 ^ 2)
x = root (452)
x = 2raiz (113)
question 6
x = root (8 ^ 2 + 15 ^ 2)
x = 17
question 7
x = root (12 ^ 2 + 12 ^ 2)
x = root (2 * (12 ^ 2))
x = 12raiz (2)
question 8
x = root (18 ^ 2 - 9 ^ 2)
x = root (243)
x = 9raiz (3)
Answer:
A. (x + 2)^2 = 31.
Step-by-step explanation:
x^2 + 4x = 27
x^2 + 4x - 27 = 0
A. (x + 2)^2 = 31
x^2 + 4x + 4 - 31 = 0
x^2 + 4x - 27 = 0
This option corresponds to the given equation, so it is correct.
B. (x + 2)^2 = 43
x^2 + 4x + 4 - 43 = 0
x^2 + 4x - 39 = 0
This does not correspond to the equation.
C. (x + 4)^2 = 31
x^2 + 8x + 16 - 31 = 0
x^2 + 8x - 15 = 0
This does not correspond to the equation.
D. (x+4)^2 = 43
x^2 + 8x + 16 - 43 = 0
x^2 + 8x - 27 = 0
This does not correspond to the equation.
So, your answer is A. (x + 2)^2 = 31.
Make sure to SUBMIT and check your PREVIOUS answers to make sure they are right! XD Also, hope this helps!
I've answered your other question as well.
Step-by-step explanation:
Since the identity is true whether the angle x is measured in degrees, radians, gradians (indeed, anything else you care to concoct), I’ll omit the ‘degrees’ sign.
Using the binomial theorem, (a+b)3=a3+3a2b+3ab2+b3
⇒a3+b3=(a+b)3−3a2b−3ab2=(a+b)3−3(a+b)ab
Substituting a=sin2(x) and b=cos2(x), we have:
sin6(x)+cos6(x)=(sin2(x)+cos2(x))3−3(sin2(x)+cos2(x))sin2(x)cos2(x)
Using the trigonometric identity cos2(x)+sin2(x)=1, your expression simplifies to:
sin6(x)+cos6(x)=1−3sin2(x)cos2(x)
From the double angle formula for the sine function, sin(2x)=2sin(x)cos(x)⇒sin(x)cos(x)=0.5sin(2x)
Meaning the expression can be rewritten as:
sin6(x)+cos6(x)=1−0.75sin2(2x)=1−34sin2(2x)
