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2 years ago

1. Suppose you pay back $680 on a $650 loan you had for 75 days. What was your simple annual interest rate?

1 answer:

3 0

well, let's assume a year has 365 days, so 75 days is simply 75/365 of a year. We also know that your loan was $650 and you paid back $680 or namely 30 bucks extra, so the interest on those $650 is really 30 bucks.

$~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill &\$30\\ P=\textit{original amount deposited}\dotfill & \$650\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\to \frac{75}{365}\dotfill &\frac{15}{73} \end{cases} \\\\\\ 30 = (650)(\frac{r}{100})(\frac{15}{73})\implies 30=\cfrac{9750r}{7300}\implies 30=\cfrac{195r}{146}\implies 4380=195r \\\\\\ \cfrac{4380}{195}=r\implies \stackrel{\%}{22.46}\approx r$

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2 years ago

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Answer:

The required equation is 3y=36 and the tape diagram is shown below.

Step-by-step explanation:

Consider the provided information.

We need to write a multiplication equation that has a solution of 12.

3y=36

We know we have to split 36 into three equal groups, so we have to divide by 3 to show this algebraically.

3y÷3=36÷3

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Now check this with the help of tape diagram as shown.

Consider the figure 1.

This represents if 36 had to be split into three groups, how big would each group be?

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3 years ago

The Cinemania theater showed 108108108 different movies last year. Of those, 151515 movies were action movies. Based on this dat

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Answer:

Option B.

Step-by-step explanation:

It is given that,

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3 years ago

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3 years ago

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$\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)$

with $0\le t\le1$ . Then

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)$

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$=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}$

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If there happens to be a scalar function $f$ such that $\vec F=\nabla f$ , then $\vec F$ is conservative and the integral is path-independent, so we only need to worry about the value of $f$ at the path's endpoints.