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3 years ago

10

1/12 +1/23+...+1/n(n+1)= n/n+1 proof by mathematical induction I need this asap!!

1 answer:

WINSTONCH [101]3 years ago

6 0

Base case (<em>n</em> = 1):

• On the left side: 1/(1×2) = 1/2

• On the right side: 1/(1 + 1) = 1/2

Induction hypothesis: Assume the statement is true for <em>n</em> = <em>k</em> ; that is,

1/(1×2) + 1/(2×3) + … + 1/(<em>k</em> × (<em>k</em> + 1))) = <em>k</em>/(<em>k</em> + 1)

Inductive step (<em>n</em> = <em>k</em> + 1):

1/(1×2) + 1/(2×3) + … + 1/(<em>k</em> × (<em>k</em> + 1))) + 1/((<em>k</em> + 1) × (<em>k</em> + 2)))

= <em>k</em>/(<em>k</em> + 1) + 1/((<em>k</em> + 1) × (<em>k</em> + 2))

= (<em>k</em> × (<em>k</em> + 2) + 1) / ((<em>k</em> + 1) × (<em>k</em> + 2))

= (<em>k</em> ² + 2<em>k</em> + 1) / ((<em>k</em> + 1) × (<em>k</em> + 2))

= (<em>k</em> + 1)² / ((<em>k</em> + 1) × (<em>k</em> + 2))

= (<em>k</em> + 1) / (<em>k</em> + 2)

and this is what we wanted to show.

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3 years ago

The value of an investment A (in dollars) after t years is given by the function A(t) = A0ekt. If it takes 10 years for an inves

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Answer:

20 years.

Step-by-step explanation:

We have been given a formula $A(t)=A_0\cdot e^{kt}$ , which represents the value of an investment A (in dollars) after t years.

Substitute the given values:

$\$3,000=\$1,000\cdot e^{k*10}$

Let us solve for k.

$\frac{\$3,000}{\$1,000}=\frac{\$1,000\cdot e^{k*10}}{\$1,000}$

$3=e^{k*10}$

Take natural log of both sides:

$\text{ln}(3)=\text{ln}(e^{k*10})$

Using property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$\text{ln}(3)=10k\cdot\text{ln}(e)$

We know that $\text{ln}(e)=1$ , so

$\text{ln}(3)=10k\cdot 1$

$\text{ln}(3)=10k$

$\frac{\text{ln}(3)}{10}=\frac{10k}{10}$

$\frac{\text{ln}(3)}{10}=k$

$\$9,000=\$1,000\cdot e^{\frac{\text{ln}(3)}{10}*t}$

Dividing both sides by 1000, we will get:

$9=e^{\frac{\text{ln}(3)}{10}*t}$

Take natural log of both sides:

$\text{ln}(9)=\text{ln}(e^{\frac{\text{ln}(3)}{10}*t)$

$\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot\text{ln}(e)$

$\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot1$

$10*\text{ln}(9)=10*\frac{\text{ln}(3)}{10}*t$

$10\text{ln}(9)=\text{ln}(3)*t$

$10\text{ln}(3^2)=\text{ln}(3)*t$

$2\cdot 10\text{ln}(3)=\text{ln}(3)*t$

$20\text{ln}(3)=\text{ln}(3)*t$

Divide both sides by $\text{ln}(3)$ :

$\frac{20\text{ln}(3)}{\text{ln}(3)}=\frac{\text{ln}(3)*t}{\text{ln}(3)}$

$20=t$

Therefore, it will take 20 years for the investment to be $9,000.

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3 years ago

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kipiarov [429]

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3 years ago

Solve for x in the equation 2 x squared + 3 x minus 7 = x squared + 5 x + 39.

Nastasia [14]

Answer:

$\implies x=1\pm \sqrt{47}$

Step-by-step explanation:

We want to solve for x in $2x^2+3x-7=x^2+5x+39$

You need to group and combine like terms and write in standard form:

$2x^2-x^2+3x-5x-7-39=0$

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By comparing to $ax^2+bx+c=0$ , we a=1,b=-2 and c=-46

The solution can be obtained using the quadratic formula.

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We substitute the coefficients to get:

$x=\frac{--2\pm \sqrt{(-2)^2-4*1*-46} }{2*1}$

$x=\frac{2\pm \sqrt{4+4*46} }{2}$

$x=\frac{2\pm \sqrt{4*47} }{2}$

$x=\frac{2\pm2\sqrt{47} }{2}$

$\implies x=1\pm \sqrt{47}$

The last choice is correct

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3 years ago

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Answer:

See below

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<u><em>Let's find the scale factor first.</em></u>

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=> Scale Factor = $\frac{6}{4}$

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2) Reflect it across the origin

3) Translate (Move) 5 units down and then 10 units left.

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3 years ago