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STALIN [3.7K]
1 year ago
11

] Complete the explanation of how you know whether the triangles are similar. If possible, find the Indicated length. A P 2.0 1.

7 B B Q 1.2 C С R Because two pairs of angles are congruent, the triangles are similar by the SSS Triangle Similarity Theorem QR =

Mathematics
1 answer:
disa [49]1 year ago
5 0

1. The two triangles are similar because the two pairs of angles are equal, therefore the third angles of both triangles must also be equal.

We can call this the AAA similarity criterion

2. To find QR, we must use ratios between 2 known similar sides of both triangles,

\frac{2.0}{1.7}=\frac{1.2}{QR}\text{  , QR=}\frac{1.2\times1.7}{2.0}=\text{ 1.02}

So, QR =1.02

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P(k)=a^k=2 3 4 find value of a that makes this is a valid probability distribution
Vesna [10]
Sounds like you're asked to find a such that

\displaystyle\sum_{k=2}^4\mathbb P(k)=\mathbb P(2)+\mathbb P(3)+\mathbb P(4)=1

In other words, find a that satisfies

a^2+a^3+a^4=1

We can factorize this as

a^4+a^3+a^2-1=a^3(a+1)+(a-1)(a+1)=(a+1)(a^3+a-1)=0

In order that \mathbb P(k) describes a probability distribution, require that \mathbb P(k)\ge0 for all k, which means we can ignore the possibility of a=-1.

Let a=y+\dfrac xy.

a^3+a-1=\left(y+\dfrac xy\right)^3+\left(y+\dfrac xy\right)-1=0
\left(y^3+3xy+\dfrac{3x^2}y+\dfrac{x^3}{y^3}\right)+\left(y+\dfrac xy\right)-1=0

Multiply both sides by y^3.

y^6+3xy^4+3x^2y^2+x^3+y^4+xy^2-y^3=0

We want to find x\neq0 that removes the quartic and quadratic terms from the equation, i.e.

\begin{cases}3x+1=0\\3x^2+x=0\end{cases}\implies x=-\dfrac13

so the cubic above transforms to

y^6-y^3-\dfrac1{27}=0

Substitute y^3=z and we get

z^2-z-\dfrac1{27}=0\implies z=\dfrac{9+\sqrt{93}}{18}
\implies y=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}
\implies a=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}-\dfrac13\sqrt[3]{\dfrac{18}{9+\sqrt{93}}}
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