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3 years ago

An equation for a line parallel to the line with equation y= 10x + 17

Mathematics

1 answer:

azamat3 years ago

3 0

To be parallel with that line the equation has to have the same slope as the equation given, like y = 10x + 3.

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Let F⃗ =2(x+y)i⃗ +8sin(y)j⃗ .

Alik [6]

Answer:

-42

Step-by-step explanation:

The objective is to find the line integral of $F$ around the perimeter of the rectangle with corners (4,0), (4,3), (−3,3), (−3,0), traversed in that order.

We will use <em>the Green's Theorem </em>to evaluate this integral. The rectangle is presented below.

We have that

$F(x,y) = 2(x+y)i + 8j \sin y = \langle 2(x+y), 8\sin y \rangle$

Therefore,

$P(x,y) = 2(x+y) \quad \wedge \quad Q(x,y) = 8\sin y$

Let's calculate the needed partial derivatives.

$P_y = \frac{\partial P}{\partial y} (x,y) = (2(x+y))'_y = 2\\Q_x =\frac{\partial Q}{\partial x} (x,y) = (8\sin y)'_x = 0$

Thus,

$Q_x -P_y = 0 -2 = - 2$

Now, by the Green's theorem, we have

$\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA = \int \limits_{-3}^{4} \int \limits_{0}^{3} (-2)\,dy\, dx \\ \\\phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-2y) \Big|_{0}^{3} \; dx\\ \phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-6)\; dx = -6x \Big|_{-3}^{4} = -42$

4 0

3 years ago

Your class earned $120.00 Saturday afternoon by washing cars to raise money for the class trip. This is 1/4 of the money needed

Ilya [14]

Answer:

$360.00

(I hope this was helpful)

5 0

3 years ago

A zoo has a total of 8 lions and tigers. The number of tigers is one less than twice the number of lions. Write a system of equa

valina [46]

Answer:

C. 2L-1

Step-by-step explanation:

4 0

3 years ago

A/ab-bsquare + b/ab-asquare?

lord [1]

Answer:

$\dfrac{(a+b)}{ab}$

Step-by-step explanation:

The given expression is :

$\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}$

It can be solved as follows :

$\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}=\dfrac{a}{b(a-b)}+\dfrac{b}{a(b-a)}\\\\=\dfrac{a}{b(a-b)}+\dfrac{b}{-a(-b+a)}\\\\=\dfrac{1}{a-b}(\dfrac{a}{b}-\dfrac{b}{a})\\\\=\dfrac{a^2-b^2}{ab(a-b)}\\\\=\dfrac{(a-b)(a+b)}{ab(a-b)}\\\\=\dfrac{(a+b)}{ab}$

So, the solution of the given expression is equal to $\dfrac{(a+b)}{ab}$ .

7 0

3 years ago

Please answer correctly !!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!!!!!

Lena [83]

Answer:

132

Step-by-step explanation:

4 0

3 years ago