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3 years ago

What property is used to change the sum of 18+6 to the equivalent expression 6(3+1)

Mathematics

1 answer:

Fittoniya [83]3 years ago

7 0

The distributive property is used to get the equivalent expression

Step-by-step explanation:

The properties on mathematical expressions are used to simplify or elaborate the expressions.

One of the propoerties is distributive properties

The distributive property states that:

Let a, b and c be three integers then

a(b+c) = ab+ac

The property can be used both ways

In given expression

18+6

$= 6.3+ 6.1\\=6(3+1)$

The distributive property is used to get the equivalent expression

Keywords: Properties

Learn more about properties at:

#LearnwithBrainly

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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

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

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

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

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Aleks04 [339]

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Solve for ppp. -5\left(p+\dfrac35\right) = -4−5(p+ 5 3 )=−4minus, 5, left parenthesis, p, plus, start fraction, 3, divided by,

Angelina_Jolie [31]

Answer:

p = 1

Step-by-step explanation:

4 - 5(p + 3/5) = -4

Find P

4 - 5(p + 3/5) = -4

4 - 5p -15/5 = -4

4 - 5p - 3 = -4

-5p + 1 = -4

-5p = -4 - 1

-5p = -5

p = -5/-5

p = 1

Check:

4 - 5(p + 3/5) = -4

4 - 5(1 + 3/5) = -4

4 - 5 - 15/5 = -4

4 - 5 - 3 = -4

4 - 8 = -4

-4 = -4

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Mashcka [7]

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