All categories

2 years ago

Here’s the answer

Mathematics

1 answer:

VLD [36.1K]2 years ago

5 0

Answer:

thx bb

Step-by-step explanation:

You might be interested in

Write a quadratic function, in standard form, that fits the set of points. Solve it as a system of three equations. (-4, 9), (0,

Tresset [83]

Answer:

$\boxed{y=2x^2+4x-7}$

Step-by-step explanation:

Let the quadratic function be

$y=ax^2+bx+c$

We substitute $(-4,9)$ into the equation to obtain;

$9=a(-4)^2+b(-4)+c$

$\Rightarrow 9=16a-4b+c---(1)$

We substitute $(0,-7)$ to obtain;

$-7=a(0)^2+b(0)^2+c$

$\Rightarrow c=-7---(2)$

We finally substitute $(1,-1)$ to obtain;

$-1=a(1)^2+b(1)^2+c$

$\Rightarrow -1=a+b+c---(3)$

We put equation (2) into equation (1) to get;

$9=16a-4b-7$

$16a-4b=16$

$\Rightarrow 4a-b=4---(4)$

$\Rightarrow -1=a+b-7$

$\Rightarrow a+b=6---(5)$

We add equation (4) and (5) to get;

$4a+a=6+4$

$\Rightarrow 5a=10$

$\Rightarrow a=2$

We put $a=2$ into equation (5) to get;

$2+b=6$

$\Rightarrow b=6-2$

$\Rightarrow b=4$

The reqiured quadratic function is

$y=2x^2+4x-7$

7 0

3 years ago

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

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