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3 years ago

Which algebraic expression is a difference with two terms

Mathematics

2 answers:

Effectus [21]3 years ago

8 0

<h2>Answer:</h2>

The algebraic expression which is a difference of two terms is:

B. 5y-7

<h2>Step-by-step explanation:</h2>

We know that any algebraic expression is a difference of two terms if there are two terms in an expression which are separated by a subtraction sign.

3y+1

In this expression we have two terms 3y and 1 which are separated by a addition sign.

Hence, option: A is incorrect.

5y-7

In this expression we have two terms as 5y and 7 which are separated by a negative sign or a subtraction sign.

Hence, option: B is correct.

-2y+5+3

In this expression we have three terms "2y", "5" and "3"

Hence, this option is incorrect.

5(y-6)

Here we have two terms 5 and (y-6) which are separated by a multiplication sign.

Hence, option: D is incorrect.

svetoff [14.1K]3 years ago

3 0

" a difference " means subtract

5y - 7......u have 2 terms (5y and 7)...and it is the difference of 2 terms.

so ur answer is : 5y - 7

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If c is the line segment connecting (x1,y1) to (x2,y2), show that the line integral of xdy-ydx=x1y2-x2y1......this is in a chapt

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Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.

At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.

We parameterize the line segment by

$\mathbf r(t)=\langle x_1,y_1\rangle(1-t)+\langle x_2,y_2\rangle t$
$\implies\mathbf r(t)=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle$

with $0\le t\le1$ . Then we find the differential:

$\mathbf r(t)\equiv\langle x,y\rangle=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle$
$\implies\mathrm d\mathbf r\equiv\langle\mathrm dx,\mathrm dy\rangle=\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt$

with $0\le t\le1$ .

Here, the line integral is

$\displaystyle\int_{\mathcal C}x\,\mathrm dy-y\,\mathrm dx=\int_{\mathcal C}\langle-y,x\rangle\cdot\langle\mathrm dx,\mathrm dy\rangle$
$=\displaystyle\int_{t=0}^{t=1}\langle-y_1-(y_2-y_1)t,x_1+(x_2-x_1)t\rangle\cdot\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt$
$=\displaystyle\int_{t=0}^{t=1}(x_1y_2-x_2y_1)\,\mathrm dt$
$=(x_1y_2-x_2y_1)\displaystyle\int_{t=0}^{t=1}\,\mathrm dt$
$=x_1y_2-x_2y_1$

as required.

4 0

4 years ago