Mathematics, 06.04.2020 19:15 caldonjoshhsms2061

In the diagram below, the expression in each circle is the result of the sum of two rectangles connected to it. complete the diagram, writing the expressions in their simplified form.

In the diagram below, the expression in each circle is the result of the sum of two rectangles conne

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Answer from: holaadios222lol

The expression in each circle is the result of the sum of two rectangles is 2x.

Given the,

in the diagram below, the expression in each circle is the result of the sum of two rectangles connected to it.

We have to determine,

the expressions in their simplified form.

According to the question,

Expression; 5x + 2y and x + y

On solving the expression in the left rectangle.

Adding both the expression,

$\rm =5x + 2y+x + y\\\\= 6x +3y$

Expression; 5x + 4y and 4x + 5y

On solving the equation on the right rectangle,

On subtract both the expression,

$\rm= 5x + 4y -4x-5y\\\\= x-y$

Therefore,

The expression in each circle is the result of the sum of two rectangles is,

$\rm Left \ side \ of \ rectangle + \Right \ side \ of \ rectangle = x+y +x -y\\\\ \ Left \ side \ of \ rectangle + \Right \ side \ of \ rectangle = 2x$

Hence, The expression in each circle is the result of the sum of two rectangles is 2x.

For more details refer to the link given below.

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Answer from: lexhorton2002

Given:

Each circle is the result of the sum of two rectangles connected to it.

To find:

The missing expression in their simplified form.

Solution:

Left side circle:

Left circle = (5x + 2y) + (x + y)

= 5x + 2y + x + y

= 6x + 7y

The expression in left side circle is 6x + 7y.

Right side rectangle:

5x + 4y = (4x + 5y) + Right rectangle

Subtract 4x + 5y from both sides.

5x + 4y - (4x + 5y) = Right rectangle

5x + 4y - 4x - 5y = Right rectangle

x - y = Right rectangle

The expression in right side rectangle is x - y.

Bottom circle:

Bottom circle = (x + y) + (x - y)

= x + y + x - y

= 2x

The expression in left side circle is 2x.

Answer from: Quest

The answer would be a

Answer from: Quest

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.

$\bf \textit{hyperbolas, vertical traverse axis } \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} [-0.35em] \rule{34em}{0.25pt} \begin{cases} h=0\\ k=0\\ a=\sqrt{5}\\ c=3 \end{cases}\implies c=\sqrt{a^2+b^2}\implies 3=\sqrt{(\sqrt{5})^2+b^2}\implies 3^2=5+b^2$

$\bf 3^2-5=b^2\implies 9-5=b^2\implies 4=b^2\implies \sqrt{4}=b\implies \boxed{2=b} [-0.35em] ~\dotfill \cfrac{(y-0)^2}{(\sqrt{5})^2}-\cfrac{(x-0)^2}{2^2}=1\implies \cfrac{y^2}{5}-\cfrac{x^2}{4}=1$

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