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german
3 years ago
9

What stocks y’all investing in? looking for some short term and maybe long term

Mathematics
1 answer:
Vadim26 [7]3 years ago
3 0

Answer:

RBLX

Step-by-step explanation:

its keeping steady at around 100$

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Find the midpoint of the segment whose endpoints are (-6,-8) and (3,6) show your work!
pogonyaev

Answer:

-1.5, -1

Step-by-step explanation:

midpoint formula

(x1 + x2 / 2) , (y1 + y2/ 2)

-6 + 3 / 2 ,   -8  + 6 / 2

3 0
3 years ago
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( PLEASE HELP )Use the table to write a proportion .
Margaret [11]
8/6 and then m/16 is the correct answer
7 0
3 years ago
Prove that for any natural n the number 3^(n+4)−3^n is divisible by 16.
gtnhenbr [62]

Answer:

Proved (See Explanation)

Step-by-step explanation:

Show that 3ⁿ⁺⁴ - 3ⁿ is divisible by 16.

This is done as follows

\frac{3^{n+4} - 3^n}{16}

From laws of indices;

aᵐ⁺ⁿ = aᵐ * aⁿ.

So, 3ⁿ⁺⁴ can be written as 3ⁿ * 3⁴.

\frac{3^{n+4} - 3^n}{16} becomes

\frac{3^n * 3^4 - 3^n}{16}

Factorize

\frac{3^n(3^4 - 1)}{16}

\frac{3^n(81 - 1)}{16}

\frac{3^n(80)}{16}

3ⁿ * 5

5(3ⁿ)

<em>The expression can not be further simplified.</em>

<em>However, we can conclude that when 3ⁿ⁺⁴ - 3ⁿ is divisible by 16, because 5(3ⁿ) is a natural whole number as long as n is a natural whole number.</em>

3 0
3 years ago
The sum of two numbers is 20. Their difference is 14. Find the numbers.
Mila [183]
3 and 17
17-3=14

hope that helps
5 0
4 years ago
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Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).
lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted C, which we can parameterize by the vector-valued function,

\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)

for 0\le t\le1, which has differential

\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

Then with x(t)=y(t)=z(t)=1-t, we have

\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r

=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt

Complete the square in the quadratic term of the integrand: t^2-2t+2=(t-1)^2+1=(1-t)^2+1, then in the integral we substitute u=1-t:

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt

\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du

Make another substitution of v=u^2:

\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv

Integrate by parts, taking

r=v+1\implies\mathrm dr=\mathrm dv

\mathrm ds=e^v\,\mathrm dv\implies s=e^v

\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv

\displaystyle=-(2e-1)+(e-1)=-e

So, we have by the fundamental theorem of calculus that

\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)

\implies-e=f(1,1,1)-2

\implies f(1,1,1)=2-e

3 0
3 years ago
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