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2 years ago

$#4 find the value of x. Round answer to nearest tenth please

Mathematics

1 answer:

Sonja [21]2 years ago

3 0

Answer:

12.9

Step-by-step explanation:

sin 59 = opposite/hypotenuse

opposite is x, the dimension facing the angle 59

hypotenuse is the longest side = 15

sin 59 = x/15

x = 15sin59 = 15 x 0.8572 =12.858 = 12.9 in the nearest tenth

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Find the domain of h(x)=x-3/x^3-36x

Dovator [93]

There is some ambiguity here which could be removed by using parentheses. I'm going to assume that you actually meant:

x-3

h(x) = ---------------

(x^3-36x)

To determine the domain of this function, factor the denominator:

x^3 - 36x = x(x^2 - 36) = x(x-6)(x+6)

The given function h(x) is undefined when the denominator = 0, which happens at {-6, 0, 6}.

Thus, the domain is "the set of all real numbers not equal to -6, 0 or 6."

Symbolically, the domain is (-infinity, -6) ∪ (-6, 0) ∪ (0, 6) ∪ (6, +infinity).

3 0

3 years ago

Suppose that F is an inverse square force field below, where c is a constant.

Andre45 [30]

Answer:

Forces in our Universe

Step-by-step explanation:

First of all we have,

$F(r) = \frac{cr}{|r|^{3} }$

and,

$r = xi+yj+zk$

We need to define a function that allows us to find said change based on r, so one of the functions that shows that change is,

$f(r) = - c /|r|$

That is,

$\nabla f = F$

For this case F is a conservative field and the line integral is independent of the path. Thus, defining $P_{1} = (x_{1}, y_{1}, z_{1})$ and $P_{2} = (x_{2}, y_{2}, z_{2})$ . So the amount of work on the movement of the object from P1 to P2 is,

$W=\int\limits_c {F} \, dr$

$W= f(P_{1}-P_{2})$

$W= \frac{c}{(x_{2}^2+ y_{2}^2+ z_{2}^2)^{1/2} } -\frac{c}{(x_{1}^2+ y_{1}^2+ z_{1}^2)^{1/2}}$

$W= c(\frac{1}{d1}-\frac{1}{d2} )$

2) The gravitational force field is given by,

$F(r) =-\frac{mMGr}{ |r|^3}$

The maximum distance from the earth to the sun is $1.52*10 ^ 8$ km and the minimum distance is $1.47*10 ^ 8$ km. The mass values of the bodies are given by m = $5.97*10 ^ {24}$ kg, M = $1.99 *19 ^ {30}$ kg and the constant G is $6.67 * 10 ^{ -11 }$ $\frac{Nm ^ 2}{kg^2}$