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alisha [4.7K]
3 years ago
10

AABC = AGHL Complete the congruence statement AB = ​

Mathematics
1 answer:
sweet [91]3 years ago
4 0

Answer:

AB = AH

Step-by-step explanation:

Hope This Helps!!

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Will award points- im currently using cos, sin and tan
MAVERICK [17]

Answer:

w = 9.82

x = 66.73

Step-by-step explanation:

By applying tangent rule in triangle ABD,

tan(81°) = \frac{\text{Opposite side}}{\text{Adjacent side}}

6.31 = \frac{62}{w}

w = \frac{62}{6.31}

w = 9.82

Similarly,

tan(39°) = \frac{\text{Opposite side}}{\text{Adjacent side}}

0.81 = \frac{62}{(x+w)}

0.81 = \frac{62}{9.83+x}

9.82 + x = \frac{62}{0.81}

x = 76.56 - 9.82

x = 66.74

3 0
3 years ago
Find the coordinates of
alexira [117]

Answer:(6,-2)

Step-by-step explanation:

M is the midpoint of AB then ,

M = \frac{A+B}{2}

5=\frac{x+4}{2}     ,     -2=\frac{y-2}{2}

10=x+4  ,      -4=y-2

x=6          ,      y= -2

so, B= (6,-2)

3 0
3 years ago
Determine whether the sequences converge.
Alik [6]
a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}

Notice that

\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}

So as n\to\infty you have a_n\to0. Clearly a_n must converge.

The second sequence requires a bit more work.

\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then a_n will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When n=2, you have

a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1

Assume a_k\ge a_{k-1}, i.e. that a_k=\sqrt{2a_{k-1}}\ge a_{k-1}. Then for n=k+1, you have

a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k

which suggests that for all n, you have a_n\ge a_{n-1}, so the sequence is increasing monotonically.

Next, based on the fact that both a_1=\sqrt2=2^{1/2} and a_2=2^{3/4}, a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}
a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}

and so on. We're getting an inkling that the explicit closed form for the sequence may be a_n=2^{(2^n-1)/2^n}, but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, a_1=2^{1/2}. Let's assume this is the case for n=k, i.e. that a_k. Now for n=k+1, we have

a_{k+1}=\sqrt{2a_k}

and so by induction, it follows that a_n for all n\ge1.

Therefore the second sequence must also converge (to 2).
4 0
3 years ago
Jillian’s school is selling tickets for a play. The tickets cost $10.50 for adults and $3.75 for students. The ticket sales for
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168 adult tickets were sold
4 0
3 years ago
Read 2 more answers
Help if good at math
Mariana [72]
Question 2 would be 6+3+n=12

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3 years ago
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