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3 years ago

10

If an investor wants to earn $250 simple interest by investing $600 at 4.5% annual interest, how many years will it take for her

to do this (to the nearest year)?

1 answer:

mixer [17]3 years ago

5 0

Answer:

5

Step-by-step explanation:

It will take her 5 years to earn that much money

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Which expression is equivalent to 4/6 of 36 ?

Vera_Pavlovna [14]

Answer:

First, find the answer for 36 ÷ 4. Which is 9.

Now calculate each expression until you get the answer of 9.

4 ÷ 36 = 4 * 1/36 = 4/36 = 1/9

36 ÷ 1/4 = 36 * 4 = 144

1/36 * 1/4 = 1/144

36 * 1/4 = 36/4 = 9

So, the last option is the answer. D. 36 * 1/4.

4 0

2 years ago

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With a 20% off coupon, you can get a pair of shoes for $72. What is the original cost of the

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Answer:

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Step-by-step explanation:

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7 0

3 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

Pam can run four laps around the track in 330 sec. Patrick can run two laps around the track in 180 sec. which person. Runs the

Aleonysh [2.5K]

Patrick because he can run 1 lap in 90 seconds

3 0

4 years ago

(Partially Done!) Verify the identity and show your work.

vladimir1956 [14]

Throughout all of these steps I'm only going to alter the left hand side (LHS). I am NOT going to change the right hand side (RHS) at all.

Before I change the LHS of the original equation, let's focus on the given identity
cot^2(x) + 1 = csc^2(x)

Since we know it's an identity, we can subtract 1 from both sides and the identity would still hold true
cot^2(x) + 1 = csc^2(x)
cot^2(x) + 1-1 = csc^2(x)-1
cot^2(x) + 0 = csc^2(x)-1
cot^2(x) = csc^2(x)-1

So we'll use the identity cot^2(x) = csc^2(x)-1

---------------------------------------------

Now onto the main equation given

cot^2(x) + csc^2(x) = 2csc^2(x) - 1
cot^2(x) + csc^2(x) = 2csc^2(x) - 1 .... note the term in bold
csc^2(x)-1 + csc^2(x) = 2csc^2(x) - 1 .... note the terms in bold
[ csc^2(x) + csc^2(x) ] - 1 = 2csc^2(x) - 1
[ 2csc^2(x) ] - 1 = 2csc^2(x) - 1
2csc^2(x) - 1 = 2csc^2(x) - 1

The bold terms indicate how the replacements occur.

So the original equation has been proven to be an identity because the LHS has been altered to transform into the RHS

7 0

3 years ago