If the CPI increased by 8.3%, therefore this means that
there was a fractional increase of 0.083. So amount increased is:
CPI increase = 141 * 0.083 = 11.703
Therefore the end of year CPI is the sum of the original and
the increase:
end of year CPI = 141 + 11.703 = 152.703
Answer:
152.703
Answer:
- 2 1/4 pounds of blueberries
Step-by-step explanation:
Blueberries - b, raspberries - x
<u>Equations as per question:</u>
<u>Eliminate b and solve for x:</u>
- 1/5x = x - 1 4/5
- x - 1/5x = 1 4/5
- 4/5x = 9/5
- x = 9/5 : 4/5
- x = 9/5 *5/4
- x = 9/4
- x = 2 1/4
Answer:
3/5
Step-by-step explanation:
9/10 and 2/3 can cross cancel
3 goes into 9, 3 times
2 goes into 10, 5 times
they both go into themselves once
our new fractions are 3/5 and 1/1 which equals 3/5
6x216 multiply them and u get 1296 then calculate the square root and your answer is 36
let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B6%7D%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%20side%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B6%5E2-5%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B36-25%7D%5Cimplies%20%5Cpm%20%5Csqrt%7B11%7D%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
