Answer:
25^9
Step-by-step explanation:
(25^3)^3 = 25^(3 * 3) = 25^9
The required fraction of a circle of 30 degree is
A circle consists of 360°, so we can write the fraction.
Which can be simplified to by dividing the numerator and denominator by 30degree.
Fraction circles are a set of nine circles of various colors.
Each circle is broken into equal fractional parts and uses the same-sized whole. The circles included are one whole as well as circles divided into halves.
Since a circle measures 360 degrees, of it would be one degree, a very angle.
Fraction of a circle of 30 degree = Fraction of circle ÷ Total area of circle
Fraction of a circle of 30 degree =
Fraction of a circle of 30 degree =
Hence, The required fraction of a circle of 30 degree is .
For more information about Fraction of Circle click the link given below.
The point-slope form of the equation for a line can be written as
... y = m(x -h) +k . . . . . . . for a line with slope m through point (h, k)
Your function gives
... f'(h) = m
... f(h) = k
a) The tangent line is then
... y = 5(x -2) +3
b) The normal line will have a slope that is the negative reciprocal of that of the tangent line.
... y = (-1/5)(x -2) +3
_____
You asked for "an equation." That's what is provided above. Each can be rearranged to whatever form you like.
In standard form, the tangent line's equation is 5x -y = 7. The normal line's equation is x +5y = 17.
Solve the exponential equation 6u = 22
Answer:
u = log 22 / log 6
u = 1.725
Step-by-step explanation:
Let's start by saying:
When ever you have an exponential equation with a different base, you attack
Answer:
lg22/lg6
Step-by-step explanation:
6^u = 22
lg(6^u) = lg22
ulg6 = lg22
u = lg22/lg6
she begins the bicep curl with her elbow bent at a 15° angle below the horizontal, in the direction of the positive x-axis.
Determine the magnitude of the torque about her elbow.
(Show work)
Answer:
54.8 Nm
Step-by-step explanation:
Torque is the cross product of the radius vector and force vector:
τ = r × F
Another way to write it is the product of the radius and force magnitudes times the sine of the angle between the vectors.
τ = rF sin θ
Here, r = 0.366 meters and F = 155 Newtons. F is in the +y direction, and r is 15° below the +x axis, so the angle between the vectors is 90° − 15° = 75°.
τ = (0.366 m) (155 N) (sin 75°)
τ = 54.8 Nm
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