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3 years ago

The sales tax in your city is 4.4% and an item costs $3 before tax.How much tax would you pay on that item?

2 answers:

5 0

$ with Tax= $3 + ($3 * 4.4%)
$ with Tax= $3 + ($3 * 0.044)
$ with Tax= $3 + $0.13
$ with Tax= $3.13

Hope this helps! :)

kotykmax [81]3 years ago

5 0

You would get $0.13 cents
Hope this has helped!

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What is the equation of the line that is parallel to 3x-5y=7 and passes through the point (-10,8)

Karo-lina-s [1.5K]

Answer: y= 3/5 + 18

Step-by-step explanation: Bruh i legit learned this a hour ago at my tuition, ok so basically simply to slope intercept from ye? so it would be y=-3/5x+7

and then you would use the distance formula and ye distance formula is square root of x sub 2 - x sub 1 squared+ y sub 2 - y sub 1 squared

4 0

3 years ago

Find how many years it would take for an investment of $3700 to grow to $6000 at an annual interest rate of 6.2% compounded semi

lorasvet [3.4K]

Answer:

7 years 11 months

Step-by-step explanation:

The future value formula for the value of a principal P invested at annual rate r compounded n times yearly for t years is ...

FV = P(1 +r/n)^(nt)

For the given numbers, we want to find t:

6000 = 3700(1 +.062/2)^(2t)

Dividing by 3700 and taking the logarithm, we get ...

6000/3700 = 1.031^(2t)

log(60/37) = 2t·log(1.031)

Dividing by the coefficient of t gives ...

t = log(60/37)/(2log(1.031)) ≈ 7.92 . . . . . years

It will take about 7 years 11 months for the investment to grow to $6000.

7 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago

A carpenter has at most $250 to spend on lumber the inequality 8x+12y<250 represents the numbers x of 2 by 8 boards and the n

malfutka [58]

Answer:

The carpenter will not be able to buy 12 '2 by 8 boards' and 14 '4 by 4 boards'.

Step-by-step explanation:

Given:

Amount a carpenter can spend at most = $250

The inequality to represent the amount he can spend on each type of board is given as:

$8x+12y$

where $x$ represents '2 by 8 boards' and $y$ represents '4 by 4 boards'.

To determine whether the carpenter can buy 12 '2 by 8 boards' and 14 '4 by 4 boards'.

Solution :

In order to check whether the carpenter can buy 12 '2 by 8 boards' and 14 '4 by 4 boards' , we need to plugin the $x=12$ and $y=14$ in the given inequality and see if it satisfies the condition or not or in other words (12,14) must be a solution for the inequality.