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ivanzaharov [21]
2 years ago
7

The number of runs scored by the home team at a baseball game is represented by ( x + 10). The number of runs scored by the visi

ting team is represented by (4x - 10). Write an expression to find how many more runs the home team scored than the visiting team. Then evaluate the expression if the value of x is 6.
Mathematics
1 answer:
Zinaida [17]2 years ago
7 0

Maria participates in a sports activity and runs the following amount__________45m

Cual es la respuesta???????

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Use a single digit times a power of 10 to express the number 866,214,000,000
BartSMP [9]

Answer:

8.66214 x 10^11

Step-by-step explanation:

question may ask to round, so the answer may be simplified to 9 x 10^11

3 0
3 years ago
Rachel purchased a prepaid phone card for $30. Long distance calls cost 6 cents a minute using this card. Rachel used her card o
Amiraneli [1.4K]

Answer: her call lasted for 26 minutes.

Step-by-step explanation:

Let x represent the number of minutes for which her call lasted.

Rachel purchased a prepaid phone card for $30. Long distance calls cost 6 cents a minute using this card. Converting 6 cents to dollars, it becomes 6/100 = $0.06

This means that the cost of x minutes of long distance call is

0.06 × x = $0.06x.

If the remaining credit on her card is $28.44, it means that

0.06x + 28.44 = 30

0.06x = 30 - 28.44

0.06x = 1.56

x = 1.56/0.06

x = 26

8 0
3 years ago
Is anyone have the answer to 9,576,984.9874 x 98.84
algol [13]

Answer:

946589196.155

Step-by-step explanation:

Hope that helps

6 0
3 years ago
Read 2 more answers
How do I calculate the perimeter of this figure?
Alisiya [41]
You just add all of the numbers up.
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

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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