I had to read this a couple of times to see this was not about time, but is a right triangle problem. If the hour hand is at three, let us consider that a leg of a right triangle, we will consider the hypotenuse as the second hand.
Sketch the picture.
The hypotenuse is 9 cm. the angle is found by:
the (hour hand) is pointing directly at the 12. I expect you mean the minute hand. The angle between the minute hand and the second hand is 25/60*360 = 150 which make the angle between the hour hand and the second hand. 150 -90 =60
so the second hand and the hour hand gives us a right triangle, with a 60 degree angle and a hyp. of 9 cm.
cos 60 = x/9
9 cos 60 =x
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x = 4.5 cm
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Answer:
Original sum of money = $2246.51
Step-by-step explanation:
Interest = $96.60
Interest is compounded 6 times in a year ; n = 6
time = 1 year ; Rate of interest (r) = 4.2%
Interest = Future Value - Principal Value ...........(1)
![\text{Future Value = }Principal\cdot (1+\frac{r}{100\times n})^{n\cdot t}\\\\\text{Substituting this value in equation (1) , We get }\\\\Interest=Principal\cdot (1+\frac{r}{100\times n})^{n\cdot t}-Principal\\\\\implies 96.60=Principal[\cdot (1+\frac{4.2}{100\times 6})^{6\cdot 1}-1]\\\\\implies Principal=\$2246.51](https://tex.z-dn.net/?f=%5Ctext%7BFuture%20Value%20%3D%20%7DPrincipal%5Ccdot%20%281%2B%5Cfrac%7Br%7D%7B100%5Ctimes%20n%7D%29%5E%7Bn%5Ccdot%20t%7D%5C%5C%5C%5C%5Ctext%7BSubstituting%20this%20value%20in%20equation%20%281%29%20%2C%20We%20get%20%7D%5C%5C%5C%5CInterest%3DPrincipal%5Ccdot%20%281%2B%5Cfrac%7Br%7D%7B100%5Ctimes%20n%7D%29%5E%7Bn%5Ccdot%20t%7D-Principal%5C%5C%5C%5C%5Cimplies%2096.60%3DPrincipal%5B%5Ccdot%20%281%2B%5Cfrac%7B4.2%7D%7B100%5Ctimes%206%7D%29%5E%7B6%5Ccdot%201%7D-1%5D%5C%5C%5C%5C%5Cimplies%20Principal%3D%5C%242246.51)
Hence, the original sum of money borrowed = $2246.51
<span>Let
CP ------> cost price
SP ------> Selling price
we know that
SP= CP + 0.45CP
Mark up = 0.45CP
Ratio of Mark up to Selling price-------> 0.45CP/(CP + 0.45CP)
= 0.45/(1+0.45)-------> </span><span>0.45/1.45=0.3103
</span><span>0.3103 multiplied by 100 = 31.03%
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the answer is
31.03%
Answer:
The value of the printer on the first year was $ 23,750.00. On the second year it was $ 22,562.5. On the third year it was $ 21,434.38.
Step-by-step explanation:
Since the printer depreciates at a rate of 5% per year, I believe the stated equation is miss typed. Therefore I'll answer this with the correct equation that would represent that setting:
In the first year the value of the printer is:
On the second year the value of the printer is:
On the third year the value of the printer is:
The value of the printer on the first year was $ 23,750.00. On the second year it was $ 22,562.5. On the third year it was $ 21,434.38.
Answer:
The solution to f(x) = t(x) is x = 2010
Option 3 is true.
Step-by-step explanation:
The first-year , second-year , and third-year enrollment values for a technical school are shown in the table below.
Year (x) First Year f(x) Second Year s(x) Third Year t(x)
2009 785 756 756
2010 740 785 740
2011 690 710 781
2012 732 732 710
2013 781 755 800
Now we will check each option.
Option 1: The solution to f(x) = s(x) is x = 2,009
In year 2009, f(x)=s(x)
But 785≠756
Thus, False
Option 2: The solution to f(x) = s(x) is x = 785
x represents year, but 785 it no year
Thus, False
Option 3: The solution to f(x) = t(x) is x = 2010
In year 2010, f(x)=t(x)=740
But 740=740
Thus, True
Option 4: The solution to f(x) = t(x) is x =740
x represents year, but 740 it no year
Thus, False
