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3 years ago

What is the relationship between sin(90 degrees) and sin(-90 degrees)

Mathematics

1 answer:

Volgvan3 years ago

4 0

Answer:

$sin(90\°)=-sin(-90\°)$

Step-by-step explanation:

we know that

$sin(90\°)=1$

$sin(-90\°)=-1$

therefore

$sin(90\°)=-sin(-90\°)$

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Hannah is giving her sister a softball for her birthday, and wants to put it in a cubic gift box to preserve the surprise. She w

Inga [223]

Answer:

B. E(s) = s^3 - 29.

Step-by-step explanation:

The amount of empty space $E(s)$ inside the cubic box is equal to the volume of the box minus the volume of the softball.

If the side length of the cubic box is $s$ , and the volume of the softball is $29in^3$ , then the amount of empty space $E(s)$ inside is

$E(s) = s^3-29$ ,

where $s^3$ is the volume of the cubic box.

Since it matches the equation we got above, choice B stands correct.

4 0

3 years ago

Hi! Please help me solve this:

Natasha2012 [34]

Answer

by 20%????

Step-by-step explanation:

8 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

alisha [4.7K]

Answer: C

Step-by-step explanation:

When you are finding the area, the formula is A= Pi R^2.

The radius is half of the diameter. To get a diameter of 6, you have to multiply 3 by 2.

6 0

3 years ago

Read 2 more answers

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

6 0

3 years ago

Two more than than 4 times a number is -18

wolverine [178]

-4..................

7 0

3 years ago