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Xelga [282]
3 years ago
12

The dimensions of the base of Box 1 are x by 3x.

Mathematics
1 answer:
Sveta_85 [38]3 years ago
5 0

Answer:

3x^2

Step-by-step explanation:

x * x  = x^2

3 * x^2 = 3x^2

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In parallelogram DEFG, DH = x + 1, HF = 3y, G H = 3 x − 4 , GH = 3x - 4, and HE = 5y + 1. Find the values of x and y. The diagra
timofeeve [1]

keeping in mind that in a parallelogram the diagonals bisect each other, namely cut each other into two equal halves.  Check the picture below.

\stackrel{GH}{3x-4}~~ = ~~\stackrel{HE}{5y+1}\implies 3x=5y+5\implies x=\cfrac{5y+5}{3} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{DH}{x+1}~~ = ~~\stackrel{HF}{3y}\implies \stackrel{\textit{substituting "x" in the equation}}{\cfrac{5y+5}{3}+1~~ = ~~3y}

\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{3\left( \cfrac{5y+5}{3}+1 \right)=3(3y)}\implies 5y+5+3=9y\implies 5+3=4y\implies 8=4y \\\\\\ \cfrac{8}{4}=y\implies \boxed{2=y}~\hfill x=\cfrac{5y+5}{3}\implies x=\cfrac{5(2)+5}{3}\implies \boxed{x=5}

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3 years ago
It would take 150 minutes to fill a swimming pool using the water from 5 taps.
mina [271]

Answer: points for points

Step-by-step explanation:

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5n + 3n<br> simplify each expression
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Answer: 8n

Step-by-step explanation: 5n + 3n

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PLEASE ANSWER THEMM!!!!
natita [175]

Answer:

a.) b⁷÷b⁴

= <u>b⁷</u>

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= b^7-4

= b³

b.) <u>x </u><u> </u><u>×</u><u> </u><u> </u><u>x⁵</u>

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5 0
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
4 years ago
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