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3 years ago

The graph shows a car's value as a function of its age.

Mathematics

2 answers:

Fudgin [204]3 years ago

8 0

The given graph shows a car's value as a function of its age.

Along x axis, these values shows the age of car in years and along y axis, these values shows the value of car after the number of years.

We have to find the value of car in 2 years.

Observe the graph below.

At 2, the value of car is $10,000.

otez555 [7]3 years ago

3 0

The car was worth $10,000.

Hope this helps!

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Answer:

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3 years ago

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2 years ago

A radio station advertises a contest with ten cash prizes totaling $5510. There is to be a $100 difference between each successi

My name is Ann [436]

Answer:

option C

c. least: $101

greatest: $1001

Step-by-step explanation:

A radio station advertises a contest with ten cash prizes totaling $5510. There is to be a $100 difference between each successive prize.

Sum of 10 prizes = 5510

100 is the difference. there is a common difference d=100

So its a arithmetic sequence

the sum formula for arithmetic sequence is

$S_n = \frac{n}{2}(2a_1 +(n-1)d)$

Sn = 5510, n=10 n d= 100 we need to find out first term a1

$5510 = \frac{10}{2}(2a_1 +(10-1)100)$

5510 = 5 (2a1 + 900)

5510 = 10a1 + 4500

Subtract 4500 on both sides

1010= 10a1

divide by 10 on both sides

a1 = 101

so first term that is least term is 101

To find out greatest term we use formula

$a_n = a_1 + (n-1) d$

a(10) = 101 + (10-1)100

= 101 + 900= 1001

greatest is 1001

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3 years ago

Make up a story about 10 things in your house

Lunna [17]

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3 years ago

Read 2 more answers

Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0

natita [175]

Answer:

$\frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C$

Step-by-step explanation:

Given differential equation,

$(x^2 + y^2) dx + (x^2 - xy) dy = 0$

$\implies \frac{dy}{dx}=-\frac{x^2 + y^2}{x^2 - xy}----(1)$

Let y = vx

Differentiating with respect to x,

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

From equation (1),

$v+x\frac{dv}{dx}=-\frac{x^2 + (vx)^2}{x^2 - x(vx)}$

$v+x\frac{dv}{dx}=-\frac{x^2 + v^2x^2}{x^2 - vx^2}$

$v+x\frac{dv}{dx}=-\frac{1 + v^2}{1 - v}$

$v+x\frac{dv}{dx}=\frac{1 + v^2}{v-1}$

$x\frac{dv}{dx}=\frac{1 + v^2}{v-1}-v$

$x\frac{dv}{dx}=\frac{1 + v^2-v^2+v}{v-1}$

$x\frac{dv}{dx}=\frac{v+1}{v-1}$

$\frac{v-1}{v+1}dv=\frac{1}{x}dx$

Integrating both sides,

$\int{\frac{v-1}{v+1}}dv=\int{\frac{1}{x}}dx$

$\int{\frac{v-1+1-1}{v+1}}dv=lnx + C$

$\int{1-\frac{2}{v+1}}dv=lnx + C$

$v-2ln(v+1)=lnx+C$

Now, y = vx ⇒ v = y/x

$\implies \frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C$

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3 years ago