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3 years ago

Solve for x if the equation is 2/5(x+1)=g,

2 answers:

8 0

In order to get x, one needs to simply rearrange the whole equation to solve for x.
$\frac{2}{5}(x+1)=g$

Multiply the two sides of the equation by 5/2.
$\frac{5}{2} [\frac{2}{5} (x+1)=g]$

Distributing the 5/2 and simplifying, the resulting equation would be:
$\frac{10}{10} (x +1)= \frac{5}{2}g$
$(x +1)= \frac{5}{2}g$

Simplifying the equation and subtracting 1 from each side of the equation the final equation for solving x is:
$x = \frac{5}{2} g - 1$

DochEvi [55]3 years ago

4 0

$\dfrac{2}{5}(x+1)=g\ \ \ |multiply\ both\ sides\ by\ \dfrac{5}{2}\\\\x+1=\dfrac{5}{2}g\ \ \ |subtract\ 1\ from\ both\ sides\\\\\boxed{x=\frac{5}{2}g-1}\to\boxed{x=\frac{5g-2}{2}}$

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Katena32 [7]

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DochEvi [55]

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3 years ago

What are the next two terms for the following sequence 1 , 8 , 27 , 64

mixas84 [53]

Answer:

125 and 216

Step-by-step explanation:

Here,

a₁ = (1)³ = 1

a₂ = (2)³ = 8

a₃ = (3)³ = 27

a₄ = (4)³ = 64

So,

a₅ = (5)³ = 125

a₆ = (6)³ 216

Thus, The next two terms for the following sequence 1 , 8 , 27 , 64 is 125 and 216

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3 years ago

How to write 12,375,003 in expanded form

icang [17]

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3 years ago

Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p

Juli2301 [7.4K]

I'm reading this as

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy$

with $\nabla f=(2xe^{-y},2y-x^2e^{-y})$ .

The value of the integral will be independent of the path if we can find a function $f(x,y)$ that satisfies the gradient equation above.

You have

$\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}$

Integrate $\dfrac{\partial f}{\partial x}$ with respect to $x$ . You get

$\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx$
$f=x^2e^{-y}+g(y)$

Differentiate with respect to $y$ . You get

$\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]$
$2y-x^2e^{-y}=-x^2e^{-y}+g'(y)$
$2y=g'(y)$

Integrate both sides with respect to $y$ to arrive at

$\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy$
$y^2=g(y)+C$
$g(y)=y^2+C$

So you have

$f(x,y)=x^2e^{-y}+y^2+C$

The gradient is continuous for all $x,y$ , so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e$

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3 years ago