$\bf \begin{array}{llll}
\cfrac{(x-{{ h}})^2}{{{ a}}^2}+\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1\\\\ \cfrac{(x-{{ h}})^2}{{{ b}}^2}+\cfrac{(y-{{ k}})^2}{{{ a}}^2}=1
\end{array}\quad 
\begin{cases}
\textit{major axis}=a+a\\
a=\textit{the larger denominator}\\
b=\textit{smaller denominator}
\end{cases}\\\\
-----------------------------\\\\
\cfrac{(x-7)^2}{4}+\cfrac{(y+3)^2}{16}=1\implies \cfrac{(x-7)^2}{2^2}+\cfrac{(y+3)^2}{4^2}=1\quad 
\begin{cases}
a=4\\
b=2
\end{cases}$

6 0

2 years ago

Arnetta’s monthly gross pay is $4,200. Federal withholding is 16.05% of her pay. Her other deductions total $321.30. What is her

Westkost [7]

Answer:

Step-by-step explanation:

8 0

2 years ago

Lamont works for an electric utility and is installing a 40 foot utility pole as shown. Lamont drills the anchor of a support ca

PIT_PIT [208]

Answer:

Approximately 43 feet of minimum cable is needed.

Step-by-step explanation:

This problem can be solved using Trigonometry and Pythagorean theorem. Pythagorean theorem applies on right-triangles (which are known to have one 90° angle). The theorem states that the square of the Hypotenuse is obtained by the squared sum of the other two sides of the triangle (i.e the two sides forming the 90° angle - with the hypotenuse side being across it as:

$h^2=a^2+b^2$ Eqn. (1)

where

$h$ is the hypotenuse

$a$ is a side

$b$ is a side

Now in this case, the utility pole must be perpendicular to the ground and the anchor being parallel to the ground, and a 90° angle formed between them. Conclusively the cable length will be represented by the hypotenuse in a right triangle. So here we have $a=40ft$ and $b=15ft$ . Plugging in Eqn.(1) and solving for $h$ we have:

$h^2=40^2+15^2\\h=\sqrt{40^2+15^2} \\h=\sqrt{1600+225}\\ h=\sqrt{1825}\\ h=5\sqrt{73}\\ h=42.72ft$

So we conclude that the minimum length of cable needed by Lamont is

≈43 feet (rounded up).

8 0

3 years ago