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2 years ago

if danny jogged 8x-21 miles and julie jogged 24x-63 mike’s, how many times longer was julie’s travel distance than danny’s?

2 answers:

6 0

Julie traveled 3 times more miles than Danny.
This is because 24x is 3 times more than 8x and -63 is 3 times more than -21.

navik [9.2K]2 years ago

5 0

I believe it is 3x 42

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F=-GMn/D^2 solve for M

notka56 [123]

M=FD^2/(-Gn)
here is the solution

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3 years ago

Suppose F⃗ (x,y)=(x+3)i⃗ +(6y+3)j⃗ . Use the fundamental theorem of line integrals to calculate the following.

Scorpion4ik [409]

In order to use the fundamental theorem of line integrals, you need to find a scalar potential function - that is, a scalar function f(x, y) for which

grad f(x, y) = F(x, y)

This amounts to solving for f such that

∂f/dx = x + 3

∂f/∂y = 6y + 3

Integrating both sides of the first equation with respect to x gives

f = 1/2 x ^2 + 3x + g(y)

Differentiating with respect to y gives

∂f/∂y = dg/dy = 6y + 3

Solving for g gives

g = ∫ (6y + 3) dy = 3y ^2 + 3y + C

and hence

f(x, y) = 1/2 x ^2 + 3x + 3y ^2 + 3y + C

(a) By the fundamental theorem, the integral of F along any path starting at the point P (1, 0) and ending at Q (3, 3) is

∫ F(x, y) • dr = f (3, 3) - f (1, 0) = 99/2 - 7/2 = 46

(b) Now we're talking about a closed path, so the integral is simply 0. We can verify this by checking the integral over the origin-containing paths:

• From the origin to P :

∫ F(x, y) • dr = f (1, 0) - f (0, 0) = 7/2 - 0 = 7/2

• From Q back to the origin:

∫ F(x, y) • dr = f (0, 0) - f (3, 3) = 0 - 99/2 = -99/2

Then the total integral is 7/2 + 46 - 99/2 = 0, as expected.

6 0

2 years ago

80 POINTS MATH

notka56 [123]

hey mate..

the position at which the tips of the scissors will be FARTHER APART is at position B.

as the angle made by the scissors at the centre is greater in position B.

3 0

2 years ago

Read 2 more answers

(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul

Aloiza [94]

Answer

(a) $F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

Step-by-step explanation:

(a) $f(t) = 20.5 + 10t + t^2 +$ δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 f(s)e^{-st} \, dt$

where a = ∞

=> $F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt$

where d(t) = δ(t)

=> $F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt$

Integrating, we have:

=> $F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3} )\left \{ {{a} \atop {0}} \right.$

Inputting the boundary conditions t = a = ∞, t = 0:

$F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $f(t) = e^{-t} + 4e^{-4t} + te^{-3t}$

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt$

$F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt$

$F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt$

Integrating, we have:

$F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.$

Inputting the boundary condition, t = a = ∞, t = 0:

$F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

3 0

3 years ago

(Trigonometry) Fill in the blank. Question in the picture. Please help

lisov135 [29]

4 feet? 20/5=4? Not too sure

8 0

3 years ago

Read 2 more answers