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2 years ago

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I WILL GIVE 5 STARS SAY THANKS AND MARK BRAINLIEST FOR RIght While training for a marathon, Jan lost 4.2% of her mass. If she lo

st 2.1 kg, what was her mass, in kilograms, before she started training?

1 answer:

daser333 [38]2 years ago

3 0

Answer:

50 kg

Step-by-step explanation:

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For what value of g does the function f(g) = g2 + 3g equal 18?

nikdorinn [45]

Answer:

B) 3

Step-by-step explanation:

f(3) = 3^2 + 3(3) = 9 + 9 = 18

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1 year ago

Three more than the quotient of a number and 8 is equal to 9

garri49 [273]

Answer:

The number is 48.

Step-by-step explanation:

x/8+3=9

x/8=9-3

x/8=6

x=6*8

x=48

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2 years ago

Find the surface area of the rectangular prism.<br> 8 m<br> 4 m<br> 1 m

Dmitry_Shevchenko [17]

Answer:

A=2(wl+hl+hw)

Step-by-step explanation:

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2 years ago

Consider a line passing through the points A(–28, –13) and B(28, 15). Type the y-value for the point C(–24, y) to ensure that po

Nostrana [21]

Step-by-step explanation:

$\frac{y_2-y_1}{x_2-x_1}=\frac{15-(-13)}{28-(-28)}\\=\frac{28}{2(28)}\\\therefore\ m=\frac{1}{2}\\\frac{y-y_1}{xl-x_1}=m]\\\frac{y+13}{x+28}=\frac{1}{2}\\2y+26=x+28\\2y=x+2\\ y=\frac{1}{2}x+1$

In order to find y for point C on AB, substitute point C in line equation if AB.

$y=\frac{1}{2}(-24)+1\\\therefore y=-12+1=11\\\therefore C(-24, -11)$

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2 years ago

For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

3 0

3 years ago