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3 years ago

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Vanessa deposited money into a bank account that earned 1.25% simple interest each year. After 1212 year, she had earned $5.00 i

n interest on the account.
If no other money was deposited into or withdrawn from the account, how much was her initial deposit

1 answer:

Reil [10]3 years ago

3 0

I=prt,,5.00=.0125*1212p,,p=5.00/.0125*1212=?? Use the calculator

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Chris has an account that pays 3.64% simple interest per year and wants to accumulate $3,080 in interest from it over the next 1

Semenov [28]

Answer:

Step-by-step explanation: A=p(1±r%)>t

A=3,080(1+3.64%)>12

=4730. 200

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2 years ago

If sin ø = 1/3, what are the values of cos ø and tan ø?

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Step-by-step explanation:

SinA=1/3=p/h

p=1,h=3,and b=?

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b=√3^2-1^2

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Now,

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Use quadrilateral ABCD to find the value of x. The figure is not drawn to scale. Use the following dimensions: m⦨ABC = 4x◦, m⦨BC

OLEGan [10]

Answer/Step-by-step explanation:

Question 1:

Interior angles of quadrilateral ABCD are given as: m<ABC = 4x, m<BCD = 3x, m<CDA = 2x, m<DAB = 3x.

Since sum of the interior angles = (n - 2)180, therefore:

$4x + 3x + 2x + 3x = (n - 2)180$

n = 4, i.e. number of sides/interior angles.

Equation for finding x would be:

$4x + 3x + 2x + 3x = (4 - 2)180$

$12x = (2)180$

$12x = 360$

$x = \frac{360}{12}$ (dividing each side by 12)

$x = 30$

Find the measures of the 4 interior angles by substituting the value of x = 30:

m<ABC = 4x

m<ABC = 4*30 = 120°

m<BCD = 3x

m<BCD = 3*30 = 90°

m<CDA = 2x

m<CDA = 2*30 = 60°

m<DAB = 3x

m<DAB = 3*30 = 90°

Question 2:

<CDA and <ADE are supplementary (angles on a straight line).

The sum of m<CDA and m<ADE equal 180°. To find m<ADE, subtract m<CDA from 180°.

m<ADE = 180° - m<CDA

m<ADE = 180° - 60° = 120°

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3 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

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2 years ago