Calculus

this is not a valid equation. I assume you mean:

2X+8Y=36

in which case the graph is a line with a slope of 1/4 and a y-intercept at 9/2

or

2X+4Y+2=36

in which case the graph is still a line with a slope of 1/2 and a y-intercept at 17/2

Let c represent cos x.

Then, we have,

4c2 - 2 = 0; whence,

4c2 = 2,

c2 = ½, and

c = ±√½ = ±½√2; that is,

cos x = ±½√2.

From this, it is evident that,

for 0 ≤ x < 360°, x = 45°, 135°, 225°, or 315°;

or, if you prefer,

for 0 ≤ x < 2π, x = ¼π, ¾π, 1¼π, or 1¾ π.

-6x+9 = 45

-6x = 45-9

-6x = 36

x = -6

It does not. It equals volume. The weight will vary depending on the substance occupying that volume.

The domain is all real numbers except when the denominator equals zero:

x2 - 4 = 0

x2 = 4

x = 2, -2

So the domain is all real numbers except 2 and -2.

-6x - 17 >= 8x + 25

-17 >= 14x + 25

-42 >= 14x

-3 >= x

Therefore x is less than or equal to -3.

x can be anything, if you don't have it in an equation form, can an be an infinite amount of solutions

38=1*2*19

58=1*2*29

highest common factor is 2

(ex-e-x)/2=8

ex-e-x=16

ex = 16-e-x

ln(ex) = ln(16-e-x)

x = ln(16-e-x)

x = ln((16ex-1)/ex)

x = ln(16ex-1) - ln(ex)

x = ln(16ex-1) - x

2x = ln(16ex-1)

e2x = eln(16e^x - 1)

e2x = 16ex-1

e2x-16ex +1 = 0

Consider ex as a whole to be a dummy variable "u". (ex=u) The above can be rewritten as:

u2-16u+1=0

u2-16u=-1

Using completing the square, we can solve this by adding 64 to both sides of the equation (the square of one half of the single-variable coefficient -16):

u2-16u+64=63

From this, we get:

(u-8)2=63

u-8=(+/-)sqrt(63)

u=sqrt(63)+8, u= 8-sqrt(63)

Since earlier we used the substitution u=ex, we must now use the u-values to solve for x.

sqrt(63)+8 = ex

ln(sqrt(63)+8)= ln(ex)

x = ln(sqrt(63)+8) ~ 2.769

8-sqrt(63) = ex

ln(8-sqrt(63)) = ln(ex)

x = ln(8-sqrt(63)) ~ -2.769

So, in the end, x~2.769 and x~-2.769. When backsubbed back into the original problem, this doesn't exactly solve the equation. Using a graphing calculator, the solution to this equation can be found to be approximately x=2.776 by graphing y=ex-e-x and y=16 and using the calculator to find the intersection of the two curves. This is pretty dang close to our calculated value, and rounding issues might account for this difference. The calculator, however, suggests that x~-2.769 is not a valid solution. This makes sense, and in fact it isn't a valid solution if you look at the graphs. This is an extraneous answer.

60,000,000,000

Sixty billion.

X^2 + 5X - 14

by inspection what two factors of - 14 add up to 5 ?

(X + 7)(X - 2)

X = - 7

X = 2

90+10=100+10+2=112

5/10

(2x^2-8x-42)/[(x^2-9)/(x^2-3x)]

=2(x^2-4x-21)/[(x+3)(x-3)/(x(x-3))]

=2(x-7)(x+3)/[(x+3)/x]

When we canceled out "(x-3)," it set up a domain restriction: x≠3. This is correct because we are not allowed to divide by 0 at any point, ever! Moving on...

=2(x-7)(x+3)(x)/(x+3)

=2(x-7)(x)

Now, x≠-3 for the same reason stated above. Moving on...

=2x(x-7)

=2x^2-14x

-1

13

As the other contributor mentioned, the standard formula for the period (T) of a simple pendulum is

T = 2*pi*sqrt(L/g)

so the period is inversely proportional to the square root of acceleration 'g'. But for practical purposes (as implied by the question) we can replace 'g' with another value, the apparent acceleration due to gravity, 'ga'. This value also takes into account the rotational speed and the distance from the center of the gravitational mass

ga = GM/r**2 - (w**2)*cos(LAT)

where:

w = angular velocity of the earth's rotation

= 2*pi/(24*3600) [rad/s]

LAT = observer's latitude (0=equator, 90deg=pole)

G = universal gravitation constant

M, r = mass, radius of planet/satellite/star we are on

Thus, the period of a simple pendulum is inversely proportional to to the sqare root of 'ga'. And this value varies with latitude, mass and distance.

So then let's answer the questions!

a) as we increase the height from sea level, the radius increases, reducing the 'ga' and this increases the period, T

b) as we go to the pole, LAT = 90deg, and cos(LAT) goes to zero. We thus INCREASE 'ga' and decrease the period

c) at the equator, LAT = 0 and cos(LAT) = 1, so we have a minimum value for 'ga', this increases the period

d) on the moon, our rotational velocity is much less (1 rev per 27.3 days) and the M is much smaller, and the r is much smaller! We are told that the 'ga' will be about 1/6 of the Earth's value, so the period will increase.

e) Here the M is colossal, so if we could withstand the heat and gravitational forces, 'ga' is much larger, so period will decrease.

0.907kg

It shouldn't relate at all.

The period of a pendulum depends only on its length,

not on how far it swings side-to-side.

-x*4*cos4x + (-1)*sin4x = -4xcos(4x) - sin(4x)

x=-7

1299