Answer:
LHS=(1-sin60)/cos60
=(1-√3÷2)/1÷2
=2(1-√3÷2)
=2-√3
RHS=(1-tan30)/(1+tan30)
={1-(1÷√3)}/{1+(1÷√3)}
={(√3-1)/√3}/{(√3+1)/√3}
=(√3-1)/(√3+1)
={(√3-1)(√3-1)}/{(√3+1)(√3-1)}
=(3-√3-√3+1)/(3-1)
=(4-2√3)/2
=2-√3
Brainliest?
Why?
We have the isosceles right triangle. The length of legs are 6. Use the Pythagorean theorem:
In general, the formula for calculating the hypotenuse in an isosceles right triangle is 
a - a leg, h - a hypotenuse
Answer:
y= CC-4.5x^2
Step-by-step explanation:
To find the general solution to the differential equation
dy + 9x dx = 0, we employ the method of separating variable as follows:
Note: { will represent the integral sign here.
Separating the variables and integrating, we have
{dy = -{9x dx
y = -(9/2)(x^2) + CC,
where CC is the given constant of integration.
This can be rearranged/simplified to yield
y= CC-4.5x^2
Answer:
the answer is the 2nd, 3rd, and 4th ones
Step-by-step explanation:
