PLS HELP I will mark BRAINLEST if correct

Mathematics

1 answer:

nydimaria [60]3 years ago

6 0

Question 1) 3
Question 3) 36in

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PLEASEEEE ANSWER ASAP

Alona [7]

Answer:

x = -1, x = 1

Step-by-step explanation:

Factor the equation. (It's a difference of squares, so we know the form will be (a-b)(a+b).)

$y = x^{2} - 1\\y = (x-1)(x+1)$

Test this by FOILing it out, if you're unsure. This is something it can be good to memorize!

Set it equal to 0.

$0 = (x+1)(x-1)$

Separate the two parenthetical expressions by the Zero Product Property.

$0 = x+1\\0 = x-1$

Solve for x!

$x=-1, x=1$

3 0

3 years ago

A mini-cone has a diameter of 3 inches and holds 10 in3 of ice cream. A mega-cone holds 49.13 in3 of ice cream. If the cones hav

marissa [1.9K]

The answer would be D

3 0

3 years ago

Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x

kykrilka [37]

Parameterize $S$ by

$\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take a normal vector to $S$ ,

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k$

which has norm

$\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v$

Then the integral of $f(x,y,z)=x^2+y^2$ over $S$ is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$