Twenty times a square of a positive integer plus 50 equals negative 40 times the square of the positive integer plus one-hundred
1 answer:
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Answer: A
Step-by-step explanation:
First, the problem is g(f(x)). You would plug in f(x) wherever you see an x in g(x). To find the domain, you take the bottom function, and set it equal to 0.
When you solve that, you get x=2. You know your domain is x≥2, but there is as asymptote at x=11. That means the graph never reaches x=11, but gets very close. You find that by setting the entire equation equal to 0 and solve from there.
The radius of the circle is 3 cm.
<u>Step-by-step explanation:</u>
Refer the attached diagram, the circle with centre O. In that given, AB is tangent given as 4 cm and distance of point from the circle, OA = 5 cm
As AB is tangent, OB (radius of circle) is perpendicular to AB (tangent at any point of circle). Therefore the angle of OBA is 90 degree.
Also, triangle OAB is a right angled triangle (refer attached diagram). By using Pythagoras theorem in right angled triangle,
Substitute the given values in the above expression, we get
Taking square root on both side, we get
Radius of the circle, OB = 3 cm
Answer:
Work done will be equal to 5.2059 lb-ft
Step-by-step explanation:
We have given force F = 10 lb
Spring is stretched to 2 in
So x = 2 in
As 1 inch = 0.0833 feet
So 2 inch = 2×0.0833 = 0.1666 feet
From hook's law we know that F = Kx , here K is spring constant and x is spring elongation
So
K = 60.024 lb/feet
Now new elongation x = 5 in
So 5 in = 5×0.0833 = 0.4165 feet
Work done is given by
So
So work done will be equal to 5.2059 lb-ft