Answer:
(x-2)^2+y^2=16
Step-by-step explanation:
If the center of the circle is (h,k) and the radius is r, then the equation of a circle is (x-h)^2+(y-k)^2=r^2 (with the Pythagorean theorem)
So, the center of this circle 0 is (2,0) and the radius is 4. Thus, the equation is (x-2)^2+y^2=16
The length of the diagonal AB of the given square is 8.660
Given that a diagram of a cube having dimensions 5×5×5 and asked to find out the length of diagonal AB(which is a body diagonal of the cube).
To calculate the body diagonal 1st we need to calculate the face diagonal via the Pythagorean theorem
Let's assume the length of the face diagonal is x
By Pythagorean theorem
Therefore, the length of the face diagonal is
Let's assume the body diagonal is y
By Pythagorean theorem
y==8.660
Therefore, The length of the diagonal AB of the given square is 8.660
Learn more about diagonal here:
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It would be more than 100 inches because in one foot is 12 inches, so take 100 x 12 = 1200 inches
where is the figures to solve.
Answer:
They have the same x-value
f(x) has the greater minimum
Step-by-step explanation:
To find the vertex of a second degree equation, in this case the minimum value, we can use the following equation:
x = -b / 2a
Remember that a second degree equation has the following form:
ax^2 + bx + c
so a = 1, b = -8 and c = 7. Now you have to substitute in the previous equation
x = - (-8) / 2(1)
x = 8 / 2
x = 4
This means that the two functions have the same x-value.
The y value of f(x) would be
f(4) = (4)^2 - 8(4) + 7
f(4) = 16 - 32 + 7
f(4) = -9
So the vertex, or minimun value of f(x) would be at the point (4, -9).
The vertex, or minimun value of g(x) is at the point (4, -4).
So f(x) has a minimum value of -9 and g(x) a minimum value of -4.