Write a verbal sentence for <br> N-8=16

A door of a lecture hall is in a parabolic shape. The door is 56 inches across at the bottom of the door and parallel to the flo

Arada [10]

Answer:

The parabolic shape of the door is represented by $y - 32 = -\frac{2}{49}\cdot x^{2}$ . (See attachment included below). Head must 15.652 inches away from the edge of the door.

Step-by-step explanation:

A parabola is represented by the following mathematical expression:

$y - k = C \cdot (x-h)^{2}$

Where:

$h$ - Horizontal component of the vertix, measured in inches.

$k$ - Vertical component of the vertix, measured in inches.

$C$ - Parabola constant, dimensionless. (Where vertix is an absolute maximum when $C < 0$ or an absolute minimum when $C > 0$ )

For the design of the door, the parabola must have an absolute maximum and x-intercepts must exist. The following information is required after considering symmetry:

$V (x,y) = (0, 32)$ (Vertix)

$A (x, y) = (-28, 0)$ (x-Intercept)

$B (x,y) = (28. 0)$ (x-Intercept)

The following equation are constructed from the definition of a parabola:

$0-32 = C \cdot (28 - 0)^{2}$

$-32 = 784\cdot C$

$C = -\frac{2}{49}$

The parabolic shape of the door is represented by $y - 32 = -\frac{2}{49}\cdot x^{2}$ . Now, the representation of the equation is included below as attachment.

At x = 0 inches and y = 22 inches, the distance from the edge of the door that head must observed to avoid being hit is:

$y -32 = -\frac{2}{49} \cdot x^{2}$

$x^{2} = -\frac{49}{2}\cdot (y-32)$

$x = \sqrt{-\frac{49}{2}\cdot (y-32) }$

If y = 22 inches, then x is:

$x = \sqrt{-\frac{49}{2}\cdot (22-32)}$

$x = \pm 7\sqrt{5}\,in$

$x \approx \pm 15.652\,in$

Head must 15.652 inches away from the edge of the door.

8 0

3 years ago

Help! Need help. Not good with this.

PIT_PIT [208]

The first answer is 7.7

the second one is 137.5

8 0

3 years ago

Read 2 more answers

4)^-5 write in a simplified fraction

erastova [34]

$\bf 4^{-5}?\qquad \textit{well, bear in mind that}\\
----------------------------\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad

\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\\ \quad \\
% negative exponential denominator
a^{{ n}} \implies \cfrac{1}{a^{- n}}
\qquad \qquad 

\cfrac{1}{a^{- n}}\implies \cfrac{1}{\frac{1}{a^{ n}}}\implies a^{{ n}}\\
----------------------------\\
thus\implies 4^{-5}\implies \cfrac{1}{4^{+5}}\implies \cfrac{1}{4^5}\implies \cfrac{1}{1024}$