If g(x)=x to the second power +3, find g(4)

Draw a pair of opposite rays

pav-90 [236]

Here is the answer to your question

3 0

3 years ago

Subtract the following polynomials. Write answer in a-b-c-d order. From 4a+7b-12c subtract 8a-9b+12d

ozzi

$(4a+7b-12c)-(8a-9b+12d)\\4a+7b-12c-8a+9b-12d$

Negative multiply negative = Positive

Negative multiply positive = negative

$-8a+4a+9b+7b-12c-12d\\-4a+16b-12c-12d$

Thus the answer is -4a + 16b - 12c - 12d.

4 0

2 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.

sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then

$\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy$

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

$\bf G(x,y) = (x, y, 3-x^2-y^2)$

with 0≤ x≤1 and 0≤ y≤1

So

$\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)$

we also have

$\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)$

and so

$\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2$

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy$

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

5 0

3 years ago

Plz help me I need help I swear I need help plz anyone help me plz help me plz I need the help from someone plz help me plz

Colt1911 [192]

2 is the y -4 is the x I believe that this is the answer

8 0

2 years ago

AnyonE wanna help?????

olga_2 [115]

Answer:

Step-by-step explanation:

a). Let the missing term = a

(a + 7) + (2x - 6) = -4x + 1

(a + 2x) + (7 - 6) = -4x + 1

(a + 2x) + 1 = -4x + 1

(a + 2x) = -4x

a = -4x - 2x

a = -6x

So the equation is,

(-6x + 7) + (2x - 6) = -4x + 1

b). (a² + p + 1) - (q + 5a + r) = 4a² -2a + 7

I have assumed p, q and r are the variables at the blank spaces.

(a² - q) + (p - 5a) + (1 - r) = 4a² - 2a + 7

By comparing both the sides of the equation,

a²- q = 4a²

q = a² - 4a²

q = -3a²

p - 5a = -2a

p = -2a + 5a

p = 3a

1 - r = 7

r = 1 - 7

r = -6

So the equation will be,

(a² + 3a + 1) - (-3a² + 5a - 6) = 4a²- 2a + 7

3 0

2 years ago