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Lerok [7]
3 years ago
6

Name a point that is located in quadrant three on a coordinate plane

Mathematics
2 answers:
mamaluj [8]3 years ago
4 0
(-4,-2) It is the bottom left side of a coordinate plane.
olya-2409 [2.1K]3 years ago
4 0
Hey there!

To start, the quadrants on a coordinate plane are numbered in counterclockwise order starting from the upper right-hand quadrant.

The upper right-hand quadrant is the quadrant that  is know is quadrant one. In this quadrant, the x values are positive, and y values are positive, otherwise known as (+x,+y).

The upper left-hand quadrant is quadrant two. The x values in this quadrant are negative, however the y values remain positive: (-x, +y)

The bottom left-hand quadrant is quadrant three. The x and y values in this quadrant are negative: (-x,-y).

Finally, the bottom right-hand quadrant is quadrant four, The x values are positive in this square and the y values are negative: (x, -y).

Knowing that quadrant three has a negative x and y value, points that would be located in this quadrant would be points like (-1, -1), (-3, -5), (-100, -90), etc. Therefore, your answer should be any point in the (-x, -y) format.

Hope this helps, and have a wonderful day! :)
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If (x + 2 ) is a factor of x3 − 6x2 + kx + 10, k =
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If a binomial x-a is a factor of a polynomial p(x), then p(a)=0.

x+2 is a factor of p(x)=x³-6x²+kx+10, so p(-2)=0.

p(-2)=0 \\
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7 0
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Suppose you saw 300 minutes of commercials this week. How many hours of commercials did you see
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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
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