Coefficient of 14x is 14
Coefficientis the number in front of the letter. For example, coeffient of 2b is 2. When there is no number in front of the letter the coefficient is 1.
Annually= per year
200 at 8% the first year would be
200 • 0.08 = 16 per year
2 years = 32 dollars in interest at the end of 2 years
Answer:
m = 60
Step-by-step explanation:
The context is very important, the 'm' represents the minutes late the parents are and from this we can eliminate some options :
- The last one because then a parent would drop their child off earlier
- The first one and the third one because it is too long
Replace m with 46x10^4 in the equation and solve.
0.036(46x10^4)^3/4 = 635.87
Round to 636
The answer is A.
<h2>✒️Area Between Curves</h2>
![\small\begin{array}{ |c|c} \hline \bold{Area\ Between\ Curves} \\ \\ \textsf{Solving for the intersection of }\rm y = x^2 + 2\textsf{ and }\\ \rm y = 4, \\ \\ \qquad \begin{aligned} \rm y_1 &=\rm y_2 \\ \rm x^2 + 2 &=\rm 4 \\ \rm x^2 &= \rm 2 \\ \rm x &=\rm \pm \sqrt{2} \end{aligned} \\ \\ \textsf{We only need the first quadrant area bounded} \\ \textsf{by the given curves so the integral for the area} \\ \textsf{would then be} \\ \\ \boldsymbol{\displaystyle \rm A = \int_{\ a}^{\ b} {\left( \begin{array}{c}\text{upper} \\ \text{function}\end{array} \right) - \left( \begin{array}{c} \text{lower} \\ \text{function} \end{array} \right)\ dx}} \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} \Big[4 - (x^2 + 2)\Big]\ dx \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} (2 - x^2)\ dx \\ \\ \rm A = \left[2x - \dfrac{x^3}{3}\right]_{0}^{\sqrt{2}} \\ \\ \rm A = 2\sqrt{2} - \dfrac{\big(\sqrt{2}\big)^3}{3} \\ \\ \rm A = 2\sqrt{2} - \dfrac{2\sqrt{2}}{3} \\ \\\red{\boxed{\begin{array}{c} \rm A = \dfrac{4\sqrt{2}}{3}\textsf{ sq. units} \\ \textsf{or} \\ \rm A \approx 1.8856\textsf{ sq. units} \end{array}}} \\\\\hline\end{array}](https://tex.z-dn.net/?f=%5Csmall%5Cbegin%7Barray%7D%7B%20%7Cc%7Cc%7D%20%5Chline%20%5Cbold%7BArea%5C%20Between%5C%20Curves%7D%20%5C%5C%20%5C%5C%20%5Ctextsf%7BSolving%20for%20the%20intersection%20of%20%7D%5Crm%20y%20%3D%20x%5E2%20%2B%202%5Ctextsf%7B%20and%20%7D%5C%5C%20%5Crm%20y%20%3D%204%2C%20%5C%5C%20%5C%5C%20%5Cqquad%20%5Cbegin%7Baligned%7D%20%5Crm%20y_1%20%26%3D%5Crm%20y_2%20%5C%5C%20%5Crm%20x%5E2%20%2B%202%20%26%3D%5Crm%204%20%5C%5C%20%5Crm%20x%5E2%20%26%3D%20%5Crm%202%20%5C%5C%20%5Crm%20x%20%26%3D%5Crm%20%5Cpm%20%5Csqrt%7B2%7D%20%5Cend%7Baligned%7D%20%5C%5C%20%5C%5C%20%5Ctextsf%7BWe%20only%20need%20the%20first%20quadrant%20area%20bounded%7D%20%5C%5C%20%5Ctextsf%7Bby%20the%20given%20curves%20so%20the%20integral%20for%20the%20area%7D%20%5C%5C%20%5Ctextsf%7Bwould%20then%20be%7D%20%5C%5C%20%5C%5C%20%5Cboldsymbol%7B%5Cdisplaystyle%20%5Crm%20A%20%3D%20%5Cint_%7B%5C%20a%7D%5E%7B%5C%20b%7D%20%7B%5Cleft%28%20%5Cbegin%7Barray%7D%7Bc%7D%5Ctext%7Bupper%7D%20%5C%5C%20%5Ctext%7Bfunction%7D%5Cend%7Barray%7D%20%5Cright%29%20-%20%5Cleft%28%20%5Cbegin%7Barray%7D%7Bc%7D%20%5Ctext%7Blower%7D%20%5C%5C%20%5Ctext%7Bfunction%7D%20%5Cend%7Barray%7D%20%5Cright%29%5C%20dx%7D%7D%20%5C%5C%20%5C%5C%20%5Cdisplaystyle%20%5Crm%20A%20%3D%20%5Cint_%7B0%7D%5E%7B%5Csqrt%7B2%7D%7D%20%5CBig%5B4%20-%20%28x%5E2%20%2B%202%29%5CBig%5D%5C%20dx%20%5C%5C%20%5C%5C%20%5Cdisplaystyle%20%5Crm%20A%20%3D%20%5Cint_%7B0%7D%5E%7B%5Csqrt%7B2%7D%7D%20%282%20-%20x%5E2%29%5C%20dx%20%5C%5C%20%5C%5C%20%5Crm%20A%20%3D%20%5Cleft%5B2x%20-%20%5Cdfrac%7Bx%5E3%7D%7B3%7D%5Cright%5D_%7B0%7D%5E%7B%5Csqrt%7B2%7D%7D%20%5C%5C%20%5C%5C%20%5Crm%20A%20%3D%202%5Csqrt%7B2%7D%20-%20%5Cdfrac%7B%5Cbig%28%5Csqrt%7B2%7D%5Cbig%29%5E3%7D%7B3%7D%20%5C%5C%20%5C%5C%20%5Crm%20A%20%3D%202%5Csqrt%7B2%7D%20-%20%5Cdfrac%7B2%5Csqrt%7B2%7D%7D%7B3%7D%20%5C%5C%20%5C%5C%5Cred%7B%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7D%20%5Crm%20A%20%3D%20%5Cdfrac%7B4%5Csqrt%7B2%7D%7D%7B3%7D%5Ctextsf%7B%20sq.%20units%7D%20%5C%5C%20%5Ctextsf%7Bor%7D%20%5C%5C%20%5Crm%20A%20%5Capprox%201.8856%5Ctextsf%7B%20sq.%20units%7D%20%5Cend%7Barray%7D%7D%7D%20%5C%5C%5C%5C%5Chline%5Cend%7Barray%7D)
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