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3 years ago

How do I do these questions?

Mathematics

1 answer:

yan [13]3 years ago

4 0

Step-by-step explanation:

The imaginary part of the number is where i is being multiplied by a constant. The real part is where there is no i present. I'll do a few examples.

26. The real part is 0, since there are no terms that don't have i. The imaginary part is 8i, since i is being multiplied by 8.

27. The real part is 7, since there is no i. The imaginary part is 3i, since i is being multiplied by 3.

31. The real part is 52, since there is no i. The imaginary part is 0i, since i is being multiplied by 0. You don't see the i because zero times i is zero.

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Two of the most popular options on a certain type of new car are automatic transmission and built-in GPS. Suppose that 90% of al

padilas [110]

Answer:

0.95 = 95% probability that the next person to purchase this car will request at least one of automatic transmission or built-in GPS

Step-by-step explanation:

We solve this question treating these probabilities as Venn sets.

I am going to say that:

Event A: Requesting automatic transmission

Event B: Requesting built-in GPS

90% of all buyers request automatic transmission

This means that $P(A) = 0.9$

82% of all buyers request built-in GPS

This means that $P(B) = 0.82$

77% of all buyers request both automatic transmission and built-in GPS.

This means that $P(A \cap B) = 0.77$

What is the probability that the next person to purchase this car will request at least one of automatic transmission or built-in GPS

This is $P(A \cup B)$ , which is given by:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$P(A \cup B) = 0.9 + 0.82 - 0.77 = 0.95$

0.95 = 95% probability that the next person to purchase this car will request at least one of automatic transmission or built-in GPS

3 0

3 years ago

Albert wants to show that tan(theta)sin(theta)+cos(theta)=sec(theta). He writes the following proof:

Nataly_w [17]

We have that
<span>tan(theta)sin(theta)+cos(theta)=sec(theta)
</span><span>[sin(theta)/cos(theta)] sin(theta)+cos(theta)=sec(theta)
</span>[sin²<span>(theta)/cos(theta)]+cos(theta)=sec(theta)

</span><span>the next step in this proof
is </span>write cos(theta)=cos²<span>(theta)/cos(theta) to find a common denominator
so

</span>[sin²(theta)/cos(theta)]+[cos²(theta)/cos(theta)]=sec(theta)<span>

</span>{[sin²(theta)+cos²(theta)]/cos(theta)}=sec(theta)<span>

remember that
</span>sin²(theta)+cos²(theta)=1
{[sin²(theta)+cos²(theta)]/cos(theta)}------------> 1/cos(theta)
and
1/cos(theta)=sec(theta)-------------> is ok

the answer is the option <span>B.)
He should write cos(theta)=cos^2(theta)/cos(theta) to find a common denominator.</span>

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

Find x for,<br> sin⁻¹ 4x + sin⁻¹ 3x = -<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20%7D%7B2%7D" id="TexFormula1" title="\

Novay_Z [31]

<h2>Explanation:</h2><h2></h2>

Let's solve this problem graphically. Here we have the following equation:

$sin^{-1}(4x) + sin^{-1}(3x) = -\frac{\pi}{2}$