Given options : Two intersecting circles are drawn with a radius in each marked. the image will be linked.
Given options : An equilateral triangle inscribed in a circle
A square inscribed in a circle
A regular pentagon inscribed in a circle
A regular hexagon inscribed in a circle.
<u>Note. When we join an intersection point of two circles and centers of the circles it would form an equilateral triangle that would be inscribe inside a common portion of both circles..</u>
Therefore, an equilateral triangle inscribed in a circle would be correct option.
She is completing an equilateral triangle inscribed in a circle.
L x W x H
so 5x5x5 = 125 cm^2 bc its a cube
Answer:
2.871*10^9
Step-by-step explanation:
add a decimal point at the end of the number ( 2,871,000,000.) then count how many places it takes to move the decimal point between the first and second number (2.8) then write the rest of the numbers higher than "0" after, giving you the 2.871
to show that the number written is in scientific notation you add 10 to the power of the amount of times you moved the decimal place. In this case, because the decimal point was moved a total of 9 times, it would be 10 to the 9th power (10^9).
You need to use a common denominator. The least common denominator of 11 and 22 is 22.
3/22 + -1/11 = 3/22 + -2/22 = 3/22 - 2/22 = 1/22
well, keeping in mind that a year has 12 months, that means that 8 months is 8/12 of a year, when Mrs Rojas pull her money out.
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\to \frac{8}{12}\dotfill &\frac{2}{3} \end{cases} \\\\\\ A=6000[1+(0.04)(\frac{2}{3})]\implies A=6000\left( \frac{77}{75} \right)\implies A=6160](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%20%5C%246000%5C%5C%20r%3Drate%5Cto%204%5C%25%5Cto%20%5Cfrac%7B4%7D%7B100%7D%5Cdotfill%20%260.04%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B8%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B2%7D%7B3%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D6000%5B1%2B%280.04%29%28%5Cfrac%7B2%7D%7B3%7D%29%5D%5Cimplies%20A%3D6000%5Cleft%28%20%5Cfrac%7B77%7D%7B75%7D%20%5Cright%29%5Cimplies%20A%3D6160)
well, she put in 6000 bucks, got back 160 extra, that's the interest earned in the 8 months.
what if she had left her money for 1 whole year, then
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\dotfill &1 \end{cases} \\\\\\ A=6000[1+(0.04)(1)]\implies A=6240](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%20%5C%246000%5C%5C%20r%3Drate%5Cto%204%5C%25%5Cto%20%5Cfrac%7B4%7D%7B100%7D%5Cdotfill%20%260.04%5C%5C%20t%3Dyears%5Cdotfill%20%261%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D6000%5B1%2B%280.04%29%281%29%5D%5Cimplies%20A%3D6240)
so had she left it in for a year, she'd have gotten 6240, namely 240 in interest, well, what fraction of a year's interest was earned? or worded differently, what fraction is 160(8 months) of 240(1 year)?
