Find the value of x.

On Claudia's birthday, her parents open a savings account and deposit SS. They also deposit $50 each

Pie

Wacom down straight to logo and and make sure he don’t get up

3 0

3 years ago

Given that / and /, which postulate or theorem can be used to show that all 3 center triangles in the Antigua and Barbuda flag a

MrRissso [65]

Given that we assume that all the bases of the triangles are parallel.

We can use AAA or Angle-Angle-Angle to prove that these triangles are similar.

Each parallel line creates the same angle when intersecting with the same side.
For example:
The bases of each triangle cross the left side of all the triangle.
Each angle made by the intersecting of the the parallel base and the side are the same.
Thus, each corresponding angle of all the triangles are congruent.
If these angles are congruent, then we have similar triangles.

4 0

3 years ago

Read 2 more answers

If <img src="https://tex.z-dn.net/?f=%20300cm%5E%7B2%7D%20" id="TexFormula1" title=" 300cm^{2} " alt=" 300cm^{2} " align="absmid

Artist 52 [7]

Check the picture below. Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm². Also, recall, the base is a square, thus, length = width = x.

$\bf \textit{volume of a rectangular prism}\\\\
V=lwh\quad 
\begin{cases}
l = length\\
w=width\\
h=height\\
-----\\
w=l=x
\end{cases}\implies V=xxh\implies \boxed{V=x^2h}\\\\
-------------------------------\\\\
\textit{surface area}\\\\
S=4xh+x^2\implies 300=4xh+x^2\implies \cfrac{300-x^2}{4x}=h
\\\\\\
\boxed{\cfrac{75}{x}-\cfrac{x}{4}=h}\\\\
-------------------------------\\\\
V=x^2\left( \cfrac{75}{x}-\cfrac{x}{4} \right)\implies V(x)=75x-\cfrac{1}{4}x^3$

so.. that'd be the V(x) for such box, now, where is the maximum point at?

$\bf V(x)=75x-\cfrac{1}{4}x^3\implies \cfrac{dV}{dx}=75-\cfrac{3}{4}x^2\implies 0=75-\cfrac{3}{4}x^2
\\\\\\
\cfrac{3}{4}x^2=75\implies 3x^2=300\implies x^2=\cfrac{300}{3}\implies x^2=100
\\\\\\
x=\pm10\impliedby \textit{is a length unit, so we can dismiss -10}\qquad \boxed{x=10}$

now, let's check if it's a maximum point at 10, by doing a first-derivative test on it. Check the second picture below.

so, the volume will then be at $\bf V(10)=75(10)-\cfrac{1}{4}(10)^3\implies V(10)=500 \ cm^3$