Really hard, buy I'll try to solve. um,since WX and YZ are perpendicular, you have to multiply since its a 90° right angle. I hoped I helped :) can I be brainliest it will make my day :)
The statement "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g" is FALSE.
Domain is the values of x in the function represented by y=f(x), for which y exists.
THe given statement is "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g".
Now we assume the 
and 
So here since g(x) is a polynomial function so it exists for all real x.
<em> </em>does not exists when 
, so the domain of f(x) is given by all real x except 6.
Now,
So now (fg)(x) does not exists when x=4, the domain of (fg)(x) consists of all real value of x except 4.
But domain of both f(x) and g(x) consists of the value x=4.
Hence the statement is not TRUE universarily.
Thus the given statement about the composition of function is FALSE.
Learn more about Domain here -
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Answer:
• Slope is the derivative of the function:
• At, x = 0
We want to find the value of cot(θ) given that sin(θ) = 3/8 and θ is an angle in a right triangle, we will get:
cot(θ) = (√55)/3
So we know that θ is an acute angle in a right triangle, and we get:
sin(θ) = 3/8
Remember that:
- sin(θ) = (opposite cathetus)/(hypotenuse)
- hypotenuse = √( (opposite cathetus)^2 + (adjacent cathetus)^2)
Then we have:
opposite cathetus = 3
hypotenuse = 8 = √(3^2 + (adjacent cathetus)^2)
Now we can solve this for the adjacent cathetus, so we get:
adjacent cathetus = √(8^2 - 3^2) = √55
And we know that:
cot(θ) = (adjacent cathetus)/(opposite cathetus)
Then we get:
cot(θ) = (√55)/3
If you want to learn more, you can read:
brainly.com/question/15345177
Answer:
Step-by-step explanation:
A man steps out of a plane at 4,000m of height above the ground.The point at which he jumps out of the plane would make a good reference point. However, if his acceleration is going to change as a result of him opening his parachute 2000m above the ground, a good reference point would be there. Keep in mind though, that his velocity at that instant would need to be known for it to be useful- otherwise the airplane reference point would be just as good with appropriate modeling....
