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aliina [53]
3 years ago
12

As soon as a new car is purchased and driven away from the dealership, it begins to lose its value, or depreciate. Alonso bought

a 1994 Plymouth Neon for $13500. One year later, the value of the car was $9000. What was the percent of decrease of the value of the car? (Note: Round answer to the nearest whole number percentage.)
Mathematics
1 answer:
timofeeve [1]3 years ago
8 0

Answer: 33.3%

Step-by-step explanation:

Cost price of the car = $13500

Value of car one year later = $9000

Decrease in value of car = $13500 - $9000 = $4500

The percent of decrease of the value of the car will then be:

= Decrease in value of car / Cost price × 100

= $4500 / $13500 × 100

= 1/3 × 100

= 33.3 %

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A point charge q1 = -2.1 ?C is located at the origin of a co-ordinate system. Another point charge q2 = 6.7 ?C is located along
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Answer 1
 By definition of the coulomb law we have that the force between two charges is given by
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 q1 = load 1 [C]
 q2 = load 2 [C]
 r = distance between charges [m]
 Substituting the values we have
 F12 = (9000000000) * ((- 2.10E-06) * (6.70E-06) / ((0.067) ^ 2))
 F12 = -28.2N
 
 answer 2

 In this case the value of r will be given by
 ROOT ((0.067) ^ 2 + (0.028) ^ 2) = 0.0726 m
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 F12 = (9000000000) * ((- 2.10E-06) * (6.70E-06) / ((0.0726) ^ 2))
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 F12x = -22.2N
 The answer is -22.2N


 answer 3
 For this case we have by sum of forces that
 F2N = F12 + F32
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 F32 = K (q2 * q3) / (r ^ 2)
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 q3 = (F32 (r ^ 2)) / (K * q2)
 substituting
 q3 = ((32.01) ((0.0726 / 2) ^ 2)) / ((9000000000) * (6.70E-06))
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4 years ago
I joined Brainly because I hate math. Especially word problems. I need help with step-by-step process to arrive at the answer. I
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3 years ago
Calculus3 - Infinite sequences and series ( URGENT!!)​
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Answer:

Limit=0

Converges

Absolutely converges

Step-by-step explanation:

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then a_{n+1}=\frac{2^{n+1} (n+1)!}{(3(n+1)+4)!}.

Let's rewrite a__{n+1} a little.

I'm going to hone in on (3(n+1)+4)! for a bit.

Distribute: (3n+3+4)!

Combine like terms (3n+7)!

I know when I have to find the limit of that ratio I'm going to have to rewrite this a little more so I'm going to do that here. Notice the factor (3n+4)! in a_n. Some of the factors of this factor will cancel with some if the factors of (3n+7)!

(3n+7)! can be rewritten as (3n+7)×(3n+6)×(3n+5)×(3n+4)!

Let's go ahead and put our ratio together.

a_{n+1}×\frac{1}{a_n}

The second factor in this just means reciprocal of {a_n}.

Insert substitutions:

\frac{2^{n+1} (n+1)!}{(3(n+1)+4)!}×\frac{(3n+4)!}{2^nn!}

Use the rewrite for (3(n+1)+4)!:

\frac{2^{n+1} (n+1)!}{(3n+7)(3n+6)(3n+5)(3n+4)!}×\frac{(3n+4)!}{2^nn!}

Let's go ahead and cancel the (3n+4)!:

\frac{2^{n+1} (n+1)!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{2^nn!}

Use 2^(n+1)=2^n × 2 with goal to cancel the 2^n factor on top and bottom:

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\frac{2(n+1)!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{n!}

Use (n+1)!=(n+1)×n! with goal to cancel the n! factor on top and bottom:

\frac{2(n+1)×n!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{n!}

\frac{2(n+1)}{(3n+7)(3n+6)(3n+5)}×\frac{1}{1}

Now since n approaches infinity and the degree of top=1 and the degree of bottom is 3 and 1<3, the limit approaches 0.

This means it absolutely converges and therefore converges.

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