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Mazyrski [523]
2 years ago
15

The following numbers divisible by 4 and 8 a) 726352​

Mathematics
1 answer:
Mariulka [41]2 years ago
5 0

Step-by-step explanation:

: last two digits are 52 and 52 is divisible by 4. Therefore, 726352 is also divisible by 4. by 8: last three digits are 352 and 352 is divisible by 8. Therefore, 726352 is also divisible by 8.

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Answer:

10.6 ounces

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Mike has a stack of 10 cards numbered 1 to 10 if he randomly chooses two cards without replacing the first one drawn what is the
Gekata [30.6K]
4 out of 10 because the amount of prime numbers from 1- 10 is 

2, 3, 5, and 7.

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What else would need to be congruent to show that ABC XYZ by ASA?
pochemuha
ASA stands for "angle side side" and we are already given two angles. That means we need a side. That eliminates two of the choices. I would assume that the answer is A, because XY and AB are in between the given angles.
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a business want to increase its number of employees by 65% to 20,460. How many employees does the company currently have?
Crazy boy [7]

Answer:

12,400

Step-by-step explanation:

Let the current employees be x

65% increase in number = 0.65x

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5 0
3 years ago
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

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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

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\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

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

surjective.

7 0
3 years ago
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