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3 years ago

9

Christian is a professional chef for a restaurant who also runs a small catering business. Christian did not have enough withhel

d from his paychecks to cover the federal taxes on the additional income and ended up having to pay an additional $500 in federal taxes at the end of the year. What would you recommend to Christian to avoid having this issue again the next year?

1 answer:

diamong [38]3 years ago

4 0

Answer:

He should pay his paychecks in time.

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The radius of the circle is increasing at a rate of 2 meters per minute and the sides of the square are increasing at a rate of

Lunna [17]

Answer:

Change in area=24 $\pi$ -48

Step-by-step explanation:

Let s will be the side of square and r will be the radius of circle.

Then two given conditions are

1)dr/dt=2 m/s

2)ds/dt=1 m/s

Area enclosed=(Area of square)-(Area of circle)

Area of square= $s^{2}$

Area of circle= $\pi r^{2}$

Area enclosed= $(\pi r^{2})-s^{2}$

dA/dt=2 $\pi$ r(dr/dt)-2s(ds/dt)

At s=24,and r=6

dA/dt=2( $\pi$ )(6)(2)-2(24)(1)

Change in area=24 $\pi$ -48

6 0

2 years ago

I need help bad!!!! Lol

bezimeni [28]

I think b is correct

4 0

3 years ago

Read 2 more answers

Given that cos (x) = 1/3, find sin (90 - x)

drek231 [11]

$\sin{(A - B)} \equiv \sin{A}\cos{B} - \cos{A}\sin{B} \\\\
\therefore \sin{(90 - x)}\\
 = \sin{(90)}\cos{(x)} - \cos{(90)}\sin{(x)} \\
= (1)\cos{(x)} - (0)\sin{(x)} \\
= \cos{x} - 0 \\
= \cos{x} \\
= \frac{1}{3}
$

3 0

3 years ago

Read 2 more answers

Determine if the given mapping phi is a homomorphism on the given groups. If so, identify its kernel and whether or not the mapp

shtirl [24]

Answer:

(a) No. (b)Yes. (c)Yes. (d)Yes.

Step-by-step explanation:

(a) If $\phi: G \longrightarrow G$ is an homomorphism, then it must hold

that $b^{-1}a^{-1}=(ab)^{-1}=\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1}$ ,

but the last statement is true if and only if G is abelian.

(b) Since G is abelian, it holds that

$\phi(a)\phi(b)=a^nb^n=(ab)^{n}=\phi(ab)$

which tells us that $\phi$ is a homorphism. The kernel of $\phi$

is the set of elements g in G such that $g^{n}=1$ . However,

$\phi$ is not necessarily 1-1 or onto, if $G=\mathbb{Z}_6$ and

n=3, we have

$kern(\phi)=\{0,2,4\} \quad \text{and} \quad\\\\Im(\phi)=\{0,3\}$

(c) If $z_1,z_2 \in \mathbb{C}^{\times}$ remeber that

$|z_1 \cdot z_2|=|z_1|\cdot|z_2|$ , which tells us that $\phi$ is a

homomorphism. In this case

$kern(\phi)=\{\quad z\in\mathbb{C} \quad | \quad |z|=1 \}$ , if we write a

complex number as $z=x+iy$ , then $|z|=x^2+y^2$ , which tells

us that $kern(\phi)$ is the unit circle. Moreover, since

$kern(\phi) \neq \{1\}$ the mapping is not 1-1, also if we take a negative

real number, it is not in the image of $\phi$ , which tells us that

$\phi$ is not surjective.

(d) Remember that $e^{ix}=\cos(x)+i\sin(x)$ , using this, it holds that

$\phi(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=\phi(x)\phi(x)$

which tells us that $\phi$ is a homomorphism. By computing we see

that $kern(\phi)=\{2 \pi n| \quad n \in \mathbb{Z} \}$ and

$Im(\phi)$ is the unit circle, hence $\phi$ is neither injective nor

surjective.

7 0

3 years ago

What is AE? Please help me

shepuryov [24]

The two triangles are similar, so will have a direct scale factor. We can find this by using the given measurements, 10÷8=1.25. So the scale factor from the smaller to the bigger triangle is 1.25. This means that 1.25×(x+6)=AE. We can solve this as we would with a regular linear equation:
1.25(x+6)=2x+6
1.25x+7.5=2x+6
1.5=0.75x
2=x
∴ AE is 2x+6=10

4 0

3 years ago