Answer:
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$4.7 × 915 = 4300.5
$4300.5 + 79.15 = 4379.65
15% of 4379.65 = 656.9475
4379.65 + 656.9745 = 5036.6245
Mrs. Jones' Bill was $5036.63
(Rounded to the nearest 1/10)
Answer:
200
Step-by-step explanation:
just look up 2% of 4 is what and you will get the same answer
Answer:
Step-by-step explanation:
The formula for determining the length of an arc is expressed as
Length of arc = θ/360 × 2πr
Where
θ is the central angle formed at the center of the circle.
r represents the radius of the circle.
π represents a constant whose value is 3.14
From the information given,
r = 21 inches
θ = 25 degrees
Length of arc formed = 25/360 × 2 × 3.14 × 21 = 9.2 inches,
<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)
#CarryOnLearning
#BrainlyForTrees
