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Novosadov [1.4K]
3 years ago
11

The sum of 5 times a number and minus −​2, plus 7 times a​ number

Mathematics
1 answer:
Olenka [21]3 years ago
6 0

Answer:

12x + 2

Step-by-step explanation:

Let the number be represented by x.

Then five times the number = 5*x

Seven times the number = 7*x

Sum of 5 times the number minus -2 = \[5*x - (-2)\] = \[5x +2\]

Adding seven times the number to this expression yields, \[5x+2+7x\]

\[= (5+7)x+2\]

\[= 12x+2\]

So the simplified expression corresponds to 12x + 2.

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Write an inequality for the sentence: The total t is less than sixteen.
RoseWind [281]
T<16 is your inequality.
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3 years ago
If the cubic container had a volume of 1,728 in³, what would its dimensions be?
sineoko [7]

Answer:

12 inches × 12 inches × 12 inches

Step-by-step explanation:

Cube volume is s^3

s^3=1728

s=\sqrt[3]{1728}

s=12

6 0
3 years ago
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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

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3 years ago
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emmasim [6.3K]

Answer: r = -5/18


Step-by-step explanation: You solve -3(1+6r)+12=14

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-18r+9=14

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r=-5/18


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monitta

Answer:

if the equation is f(x)=ax^2+bx+c

Step-by-step explanation:

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3 years ago
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