I'm going to assume that the room is a rectangle.
The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle.
You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:

Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:

Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.
The dimensions of your room are 25ft (length) by 24ft (width).
2x-5=95 due to the alternate interior angles thm so you just add 5 to 95 to get 2x=100 then divide by 2 to get x=50!
If A and B are equal:
Matrix A must be a diagonal matrix: FALSE.
We only know that A and B are equal, so they can both be non-diagonal matrices. Here's a counterexample:
![A=B=\left[\begin{array}{cc}1&2\\4&5\\7&8\end{array}\right]](https://tex.z-dn.net/?f=A%3DB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C4%265%5C%5C7%268%5Cend%7Barray%7D%5Cright%5D)
Both matrices must be square: FALSE.
We only know that A and B are equal, so they can both be non-square matrices. The previous counterexample still works
Both matrices must be the same size: TRUE
If A and B are equal, they are literally the same matrix. So, in particular, they also share the size.
For any value of i, j; aij = bij: TRUE
Assuming that there was a small typo in the question, this is also true: two matrices are equal if the correspondent entries are the same.
If he wants to loose 30 points total, subtract what he lost already, 24, from 30. He needs to lose 6 more pounds.
The nature of a graph, which has an even degree and a positive leading coefficient will be<u> up left, up right</u> position
<h3 /><h3>What is the nature of the graph of a quadratic equation?</h3>
The nature of the graphical representation of a quadratic equation with an even degree and a positive leading coefficient will give a parabola curve.
Given that we have a function f(x) = an even degree and a positive leading coefficient. i.e.
The domain of this function varies from -∞ < x < ∞ and the parabolic curve will be positioned on the upward left and upward right x-axis.
Learn more about the graph of a quadratic equation here:
brainly.com/question/9643976
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