How to solve stacked similar figures

Mathematics

1 answer:

Aleonysh [2.5K]3 years ago

4 0

Answer:

Step-by-step explanation:

You can identify similar polygons by comparing their corresponding angles and sides. As you see in the following figure, quadrilateral WXYZ is the same shape as quadrilateral ABCD, but it’s ten times larger (though not drawn to scale, of course). These quadrilaterals are therefore similar.

similar polygons: For two polygons to be similar, both of the following must be true:

Corresponding angles are congruent.

Corresponding sides are proportional.

To fully understand this definition, you have to know what corresponding angles and corresponding sides mean. (Maybe you’ve already figured this out by just looking at the figure.) Here’s the lowdown on corresponding. In the figure, if you expand ABCD to the same size as WXYZ and slide it to the right, it’d stack perfectly on top of WXYZ.

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What is the equation of the axis of symmetry for the parabola y equals start fraction one over three end fraction left parenthes

wariber [46]

The correct axis of symmetry is x = -1.

Explanation:

Our equation is
$y=\frac{1}{3}(x+1)^2-3$ .

This is in vertex form, which is
y = a(x-h)² + k, where (h, k) is the vertex.

In our equation, h corresponds with -1 and k corresponds with -3, making the vertex (-1, -3).

The axis of symmetry is the x-coordinate of the vertex; this makes the axis of symmetry for this equation x = -1.

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3 years ago

. Evaluate the series 1 + 0.1 + 0.01 + . . .

EastWind [94]

We can employ a simple repeated decimal trick:

$x=1.111\ldots$

$0.1x=0.111\ldots$

$\implies x-0.1x=1\implies0.9x=1\implies x=\dfrac1{0.9}=\dfrac{10}9$

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Alternatively, we can compute the partial sum of the series.

$\displaystyle S_n=\sum_{k=0}^n\dfrac1{10^k}$

$S_n=1+0.1+0.01+\cdots+\dfrac1{10^n}$

$0.1S_n=0.1+0.01+0.001+\cdots+\dfrac1{10^{n+1}}$

$\implies S_n-0.1S_n=0.9S_n=1-\dfrac1{10^{n+1}}$

$\implies S_n=\dfrac{10}9-\dfrac9{10^n}$

As $n\to\infty$ , the second term vanishes and we're left with $\dfrac{10}9$ . Notice that this is really just a more formal version of the earlier trick.