Answer:
Step-by-step explanation:
Earth volume is approximately V = (4/3)πr³, and
Uranus volume is approximately V = (4/3)π(4r)³, or (4/3)π*64*r³
Comparing these two results, we see that the volume of Uranus is approximately 64 times greater than the volyume of Earth.
There should be no problem in finding the value of the unknown variable "b" in the equation given in the question.The equation is solvable for finding the value of "b" because it is the only unknown variable in the single equation that is given in the question.
45 = 3b + 69
Let us reverse both sides of the equation first. then, we get
3b + 69 = 45
3b = 45 - 69
3b = - 24
b = - (24/3)
= - 8
So from the above deduction, we can easily conclude that the value of b in the given equation is -8.
m(slope)= change in y/change in x
m=-17-(-5)/3-1
m=17+5/2
m=22/2
m=11
Circumference of a circle - derivation
This page describes how to derive the formula for the circumference of a circle.
Recall that the definition of pi (π) is the circumference c of any circle divided by its diameter d. Put as an equation, pi is defined as
π
=
c
d
Rearranging this to solve for c we get
c
=
π
d
The diameter of a circle is twice its radius, so substituting 2r for d
c
=
2
π
r
If you know the area
Recall that the area of a circle is given by
area
=
π
r
2
Solving this for r
r
2
=
a
π
So
r
=
√
a
π
The circumference c of a circle is
c
=
2
π
r
<span>1. a sine curve with amplitude 2, and period 4pi radians
</span>
the general equation of the sine curve ⇒⇒ y = a sin (nθ)
where: a is the amplitude and n = 2π/perid
∵ <span>amplitude 2, and period 4pi radians
</span>
∴ y = 2 sin (θ/2)
The correct answer is option D. y = 2 sin (θ/2)
===========================================
<span>2.The period and amplitude of the function ⇒⇒ y = 5 cos 2θ
</span>
<span>comparing with y = a cos nθ
</span>
where : a is the amplitude and n = 2π/period
<span>amplitude = 5 , period = 2π/n = 2π/2 = π
</span>
The correct answer is option B. Period: pi radians: Amplitude:5
============================================================
3. tan (2π/3) = tan 120° = -√3
120° lie in the second quadrant and its reference angle = 180° - 120° = 60°
tan function in the second quadrant is negative
∴ tan 120° = - tan 60 = -√3
The correct answer is C. -sqrt3
=====================================================
4. <span>Tan 5π/6 = tan 150° = -(√3)/3
</span>
150° lies in the second quadrant and its reference angle = 180° - 150° = 30°
tan function in the second quadrant is negative
∴ tan 150° = - tan 30 = -(√3)/3
The correct answer is <span>B.-sqrt3/3</span>
