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3 years ago

15

A shirt that normally costs $30 is on sale for $21.75. What percent of the regular price is the sale price?

1 answer:

disa [49]3 years ago

8 0

First you divide sale cost which is $21.75 by the original cost of the shirt being $30.

Once you divide 21.75 into 30 dollars you'll get 0.725

Now you wanna multiply that 0.725 by 100% which will then

give you the total 72.5%

________________________

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Please help I don’t understand how to do this

horsena [70]

Answer:

DOC = 70

BOC = 110

AOD = 110

Step-by-step explanation:

AOB is supplementary to BOC and AOD, which means together they equal 180. 180-70=110, so BOC and AOD are both 110.

AOB and DOC are vertical angles to each other, so are AOD and BOC. Vertical angles are always equal to each other. So since AOB is 70, so is DOC.

Hope this makes sense:)

5 0

2 years ago

Lincoln invested $49,000 in an account paying an interest rate of 6\tfrac{1}{8}681% compounded daily. Eli invested $49,000 in a

timofeeve [1]

Answer: 15856

Step-by-step explanation: 166786.9963 - 150930.6256

= 15856.3707

= $15856

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1 year ago

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

3 0

2 years ago

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creativ13 [48]

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4 0

3 years ago

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MAXImum [283]

Yes i believe they can have the same vertex

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3 years ago

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