Lines m and p are perpendicular if the slope of line is 1, 25th what is the slope of line p

Which one of the following statements is true?

kicyunya [14]

The first derivitive is the slope
a positive 1st deriviive is positive slope or increasing
a negaitve 1st derivitive is negative slope or decreasing

the 2nd derivitive tells about the concavity
a negative second derivitive means it is concave down at that point
a positive 2nd derivitive means it is concave up at that point

so

first one is false
2nd is true
3rd is false because it could possibly be a minimum as well

answer is 2nd one

3 0

2 years ago

HI could some one please help me I will give you BRAINLEST!!!!

tatuchka [14]

Answer:

16

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

I need help so please help

arlik [135]

Answer:

M= sq root of 2 and N= sq root of 5 would be irrational

Step-by-step explanation:

These two answer choices are irrational because they cannot be shown or expressed as a fraction.

6 0

1 year ago

No Explanation Please Don't Answer

vazorg [7]

Answer:

A) $\frac{1}{2}(9+15)w$

Step-by-step explanation:

Area of a trapezoid is $\frac{1}{2}(base1+base2) *h$

$A=\frac{1}{2}(9+15)*w$

8 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

2 years ago