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4 years ago

Find "x"

1 answer:

4 0

1. −2x−4=20+6x
Step 1: Simplify both sides of the equation.
−2x−4=20+6x
−2x+−4=20+6x
−2x−4=6x+20
Step 2: Subtract 6x from both sides.
−2x−4−6x=6x+20−6x
−8x−4=20
Step 3: Add 4 to both sides.
−8x−4+4=20+4
−8x=24
Step 4: Divide both sides by -8.
−8x/−8=24/−8
x=−3
Answer:
x=−3

3. (x/4)*(4)=(9)*(4)
x=36
Answer:
x=36

4.
2(5x−1)=3x+5
Step 1: Simplify both sides of the equation.
2(5x−1)=3x+5
(2)(5x)+(2)(−1)=3x+5(Distribute)
10x+−2=3x+5
10x−2=3x+5
Step 2: Subtract 3x from both sides.
10x−2−3x=3x+5−3x
7x−2=5
Step 3: Add 2 to both sides.
7x−2+2=5+2
7x=7
Step 4: Divide both sides by 7.
7x/7=7/7
x=1
Answer:
x=1

5.
8−2(3x−4)+2x=5(x+6)+4
Step 1: Simplify both sides of the equation.
8−2(3x−4)+2x=5(x+6)+4
8+(−2)(3x)+(−2)(−4)+2x=(5)(x)+(5)(6)+4(Distribute)
8+−6x+8+2x=5x+30+4
(−6x+2x)+(8+8)=(5x)+(30+4)(Combine Like Terms)
−4x+16=5x+34
−4x+16=5x+34
Step 2: Subtract 5x from both sides.
−4x+16−5x=5x+34−5x
−9x+16=34
Step 3: Subtract 16 from both sides.
−9x+16−16=34−16
−9x=18
Step 4: Divide both sides by -9.
−9x/−9=18/−9
x=−2
Answer:
x=−2

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A system contains n atoms, each of which can only have zero or one quanta of energy. How many ways can you arrange r quanta of e

My name is Ann [436]

Answer:

$\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)} \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$

Step-by-step explanation:

If the system contains $n$ atoms, we can arrange $r$ quanta of energy in

$\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$

ways.

$\mathbf{a)}$

In this case,

$n = 2, r=1$ .

Therefore,

$\binom{n}{r} = \binom{2}{1} = \dfrac{2!}{1!(2-1)!} = \frac{2 \cdot 1}{1 \cdot 1} = 2$

which means that we can arrange 1 quanta of energy in 2 ways.

$\mathbf{b)}$

In this case,

$n = 20, r=10$ .

Therefore,

$\binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756$

which means that we can arrange 10 quanta of energy in 184 756 ways.

$\mathbf{c)}$

In this case,

$n = 2 \times 10^{23}, r = 10^{23}$ .

Therefore, we obtain that the number of ways is

$\binom{n}{r} = \binom{2\times 10^{23}}{10^{23}} = \dfrac{(2\times 10^{23})!}{(10^{23})!(2\times 10^{23} - 10^{23})!} = \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$

3 0

3 years ago

F(x) = 2(x + 1)2 – 3 Calculate the rate of change of the function between x=-1 and x=1

Zolol [24]

Answer:

The rate of change is 4

Step-by-step explanation:

The rate of change can be calculated using the formula;

f(b)-f(a)/b-a

According to this question, a = -1 and b = 1

f(a) = f(-1) = 2(-1+1)^2 -3 = -3

f(b) = f(1) = 2(1+1)^2 - 3 = 8-3 = 5

So the rate of change is;

5-(-3)/1-(-1) = (5+ 3)/(1 + 1) = 8/2 = 4

6 0

3 years ago

the diagonal length of a square window is 40 cm. find the area of the window. Btw any help is appreciated ! :)

7nadin3 [17]

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Hope this helped :)

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3 years ago

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Answer:

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3 years ago

Factor completely: 4x2 + 25x + 6

IceJOKER [234]

(x + 6)(4x + 1)
x times 4x = 4x^2
x times 1 = x
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6 times 1 = 6
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I hope this helps you! Good luck :)

7 0

3 years ago