In the 
plane, we have 
everywhere. So in the equation of the sphere, we have
which is a circle centered at (2, -10, 0) of radius 4.
In the 
plane, we have 
, which gives
But any squared real quantity is positive, so there is no intersection between the sphere and this plane.
In the 
plane, 
, so
which is a circle centered at (0, -10, 3) of radius 
.
Just multiply 4*12 and add 6.
answer =54
The simplest solution:
y - intercept of the function on the graph is (0; 6).
Only function f(x) = (4x^2 - 7x - 2)(2x - 3) has that y-intercept
f(0) = (4 · 0² - 7 · 0 - 2)(2 · 0 - 3) = (-2)(-3) = 6
Other functions have y - intercept equal (0; -6).
f(0) = (0 ... - 2)(0 + 3) = (-2)(3) = -6 or f(0) = (0 ... + 2)(0 - 3) = (2)(-3) = -6
To find x:
Subtract 10 on both sides.
10 + 2x = 90
-10 -10
Divide 2 on both sides.
2x = 80
--- ---
2 2
This would get x alone.
x = 40
Check:
10 + 2(40) = 90
10 + 80 = 90
90 = 90
The value of x is 40. The angle degree of it is 80 when you multiply 2 by 40.
Answer:
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