The figure below shows a parallelogram PQRS:

Mathematics

2 answers:

maw [93]3 years ago

8 0

Answer by Mimiwhatsup: Triangle PQS is congruent to triangle RSQ

Why is this right by Mimiwhatsup: As the PQ shows it is equal to SR

lawyer [7]3 years ago

4 0

The figure below shows a parallelogram PQRS:

A parallelogram PQRS is shown with the diagonal SQ.

The flowchart shown below shows the sequence of steps to prove the theorem: Opposite angles of a parallelogram are equal:

Which is the missing statement?

Answer - Triangle PQS is congruent to triangle RSQ

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What number is 56% of 48?

Digiron [165]

X = 56% * 48

x = 56/100 * 48

x = 28/50 * 48

x = 14/25 * 48

x = 672/25

x = 26.88

<span>56% of 48 is </span>26.88<span>.</span>

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4 years ago

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3 years ago

Which equation represents a hyperbola with a center at (0, 0), a vertex at (−48, 0), and a focus at (50, 0)?

Lerok [7]

bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.

we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.

we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.

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

$\bf \begin{cases} h=0\\ k=0\\ a=48\\ c=50 \end{cases}\implies \cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{b^2}=1 \\\\\\ c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b\implies \sqrt{50^2-48^2}=b \\\\\\ \sqrt{196}=b\implies 14=b~\hspace{3.5em}\cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{14^2}=1\implies \cfrac{x^2}{48^2}-\cfrac{y^2}{14^2}=1$