What’s the coordinates of point w

Mathematics

1 answer:

joja [24]3 years ago

4 0

Answer:

(-4 , -3)

Step-by-step explanation:

(x value, y value)

You always count the x value first, and in this case you have to go to -4.

(-4 , y value)

then, you determine the y value, which in this case is -3.

So, your answer is:

(-4 , -3).

Hope this helps!! Have a great day!! ~Carley

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$\small\begin{array}{ |c|c} \hline \bold{Area\ Between\ Curves} \\ \\ \textsf{Solving for the intersection of }\rm y = x^2 + 2\textsf{ and }\\ \rm y = 4, \\ \\ \qquad \begin{aligned} \rm y_1 &=\rm y_2 \\ \rm x^2 + 2 &=\rm 4 \\ \rm x^2 &= \rm 2 \\ \rm x &=\rm \pm \sqrt{2} \end{aligned} \\ \\ \textsf{We only need the first quadrant area bounded} \\ \textsf{by the given curves so the integral for the area} \\ \textsf{would then be} \\ \\ \boldsymbol{\displaystyle \rm A = \int_{\ a}^{\ b} {\left( \begin{array}{c}\text{upper} \\ \text{function}\end{array} \right) - \left( \begin{array}{c} \text{lower} \\ \text{function} \end{array} \right)\ dx}} \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} \Big[4 - (x^2 + 2)\Big]\ dx \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} (2 - x^2)\ dx \\ \\ \rm A = \left[2x - \dfrac{x^3}{3}\right]_{0}^{\sqrt{2}} \\ \\ \rm A = 2\sqrt{2} - \dfrac{\big(\sqrt{2}\big)^3}{3} \\ \\ \rm A = 2\sqrt{2} - \dfrac{2\sqrt{2}}{3} \\ \\\red{\boxed{\begin{array}{c} \rm A = \dfrac{4\sqrt{2}}{3}\textsf{ sq. units} \\ \textsf{or} \\ \rm A \approx 1.8856\textsf{ sq. units} \end{array}}} \\\\\hline\end{array}$