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Anna71 [15]
1 year ago
5

Can someone help me out on this geometry postulate questions? It’s urgent

Mathematics
1 answer:
Ugo [173]1 year ago
6 0

20) \overline{AC} \cong \overline{AC} by the reflexive property, so SSS

21) \overline{CD} \cong \overline{CD} by the reflexive property, so ASA

22) \overline{CE} \cong \overline{CB}, \overline{AC} \cong \overline{CD} by the definition of a midpoint and \angle 1 \cong \angle 2, so SAS

23) \overline{AC} \cong \overline{AC} by the reflexive property, so ASA

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13+39-2-10-6 power2 orders of operation
Advocard [28]
6 to the power of 2 is 6 multiplied by six which is 36
then 13 + 39 is 52 - 2 = 50- 10 = 40 - 36=  the answer 4
8 0
3 years ago
Which of the following geometric series converges?
Artist 52 [7]

All three series converge, so the answer is D.

The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.

Consider a geometric sequence with the first term <em>a</em> and common ratio |<em>r</em>| < 1. Then the <em>n</em>-th partial sum (the sum of the first <em>n</em> terms) of the sequence is

S_n=a+ar+ar^2+\cdots+ar^{n-2}+ar^{n-1}

Multiply both sides by <em>r</em> :

rS_n=ar+ar^2+ar^3+\cdots+ar^{n-1}+ar^n

Subtract the latter sum from the first, which eliminates all but the first and last terms:

S_n-rS_n=a-ar^n

Solve for S_n:

(1-r)S_n=a(1-r^n)\implies S_n=\dfrac a{1-r}-\dfrac{ar^n}{1-r}

Then as gets arbitrarily large, the term r^n will converge to 0, leaving us with

S=\displaystyle\lim_{n\to\infty}S_n=\frac a{1-r}

So the given series converge to

(I) -243/(1 + 1/9) = -2187/10

(II) -1.1/(1 + 1/10) = -1

(III) 27/(1 + 1/3) = 18

8 0
3 years ago
If the lengths of the right triangle are 7 and 8 what is the approximate hypotenuse
ANEK [815]
<h3>Answer:</h3>

c=√a²+b²

=√7²+8²

<h3> C ≈10.63015 or 10.6</h3>

Hope this helps! <3

4 0
3 years ago
Need help with number 1.
seropon [69]

Answer:

Your answer is E.

Step-by-step explanation:

8 0
3 years ago
Help me please,Thanks
Anna71 [15]
She has 7 more to pack
7 0
3 years ago
Read 2 more answers
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