Please help asap!!!!! At Sarah's cafe, customers can choose one of the following entrees: burrito, pasta, soup or taco. They can
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Well, parallel lines have the same exact slope, so hmmm what's the slope of the one that runs through <span>(0, −3) and (2, 3)?
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so, we're really looking for a line whose slope is 3, and runs through -1, -1
</span>![\bf \begin{array}{ccccccccc} &&x_1&&y_1\\ % (a,b) &&(~ -1 &,& -1~) \end{array} \\\\\\ % slope = m slope = m\implies 3 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-1)=3[x-(-1)] \\\\\\ y+1=3(x+1)](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%20-1%20%26%2C%26%20-1~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20slope%20%20%3D%20m%0Aslope%20%3D%20%20m%5Cimplies%203%0A%5C%5C%5C%5C%5C%5C%0A%25%20point-slope%20intercept%0A%5Cstackrel%7B%5Ctextit%7Bpoint-slope%20form%7D%7D%7By-%20y_1%3D%20m%28x-%20x_1%29%7D%5Cimplies%20y-%28-1%29%3D3%5Bx-%28-1%29%5D%0A%5C%5C%5C%5C%5C%5C%0Ay%2B1%3D3%28x%2B1%29)
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so, we're really looking for a line whose slope is 3, and runs through -1, -1
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(See attached formula)
Years = log (total / principal) / n*log(1 + rate/n) where n = compounding periods per year
Years = log (total / principal) / n*log(1 + rate/n)
Rate of .75% = .0625 per month
We'll say we want total = 200 and principal = 100
Years = log (200 / 100) / 12*log(1 + rate/n)
Years = log (2) / 12 * log (1 + .0625/12)
Years = <span> <span> <span> 0.3010299957 </span> </span> </span> / 12 * <span> <span> <span> 0.0022560803 </span> </span> </span> <span><span> </span> </span>
Years = <span> <span> 0.3010299957 </span> </span> / <span> <span> <span> 0.0270729635 </span>
Years = </span></span><span><span><span>11.1192110871 </span> </span> </span>
I don't think that's right
The interest rate is REALLY low. (Less than 1%)
Rate = .75 and .0075 when in a formula
Years = log (2) / 12 * log (1 + .0075/12)
Years = log (2) / 12 * log( <span>1.000625</span>)
Years = <span> <span> <span> 0.3010299957 </span> </span> </span> / 12 * <span> <span> <span> 0.0002713493 </span> </span> </span>
Years = <span><span><span> <span> 0.3010299957 </span> </span> / 0.0032561912 </span>
</span>Years = <span><span><span>92.4485010892
THAT seems a more likely answer with such a low interest rate.
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Years = log (total / principal) / n*log(1 + rate/n) where n = compounding periods per year
Years = log (total / principal) / n*log(1 + rate/n)
Rate of .75% = .0625 per month
We'll say we want total = 200 and principal = 100
Years = log (200 / 100) / 12*log(1 + rate/n)
Years = log (2) / 12 * log (1 + .0625/12)
Years = <span> <span> <span> 0.3010299957 </span> </span> </span> / 12 * <span> <span> <span> 0.0022560803 </span> </span> </span> <span><span> </span> </span>
Years = <span> <span> 0.3010299957 </span> </span> / <span> <span> <span> 0.0270729635 </span>
Years = </span></span><span><span><span>11.1192110871 </span> </span> </span>
I don't think that's right
The interest rate is REALLY low. (Less than 1%)
Rate = .75 and .0075 when in a formula
Years = log (2) / 12 * log (1 + .0075/12)
Years = log (2) / 12 * log( <span>1.000625</span>)
Years = <span> <span> <span> 0.3010299957 </span> </span> </span> / 12 * <span> <span> <span> 0.0002713493 </span> </span> </span>
Years = <span><span><span> <span> 0.3010299957 </span> </span> / 0.0032561912 </span>
</span>Years = <span><span><span>92.4485010892
THAT seems a more likely answer with such a low interest rate.
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