Help please, I have no idea how to do this and I have a test on it tmrw!

Mathematics

2 answers:

aev [14]2 years ago

6 0

Answer:

Step-by-step explanation:

Supplementary means that 2 angles add up to 180 degrees.

If ACB and ACD are supplementary then they have to add up to 180 degrees.

We know that ACD is smaller than ACB and EGF is supplementary to the smaller of the two.

So then these two are supplementary.

If these are supplementary then the only reason i can think of is that they are supplementary or not. I recently had a test on this too and i may have failed. I have given you all the information i know. Sorry. If it is multiple choce then pick a random one. You may get it correct.

arsen [322]2 years ago

5 0

Answer:

See below.

Step-by-step explanation:

We didn't do column proofs when I attended school (in the UK) so I can only show you how we would prove it in the UK.

The angles are supplementary (given).

so < ACB + < ACD = 180

< EGF + < ACD = 180

Subtracting:

< ACB - <EGF = 0

Therefore < ACB ≅ <EGF.

I hope this helps.

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4 0

1 year ago

Infinite over E n=1 4(0.5)^n-1

yan [13]

Answer:

S = 8

Step-by-step explanation:

An infinite geometric series is defined as limit of partial sum of geometric sequences. Therefore, to find the infinite sum, we have to find the partial sum first then input limit approaches infinity.

However, fortunately, the infinite geometric series has already set up for you. It’s got the formula for itself which is:

$\displaystyle \large{\lim_{n\to \infty} S_n = \dfrac{a_1}{1-r}}$

We can also write in summation notation rather S-term as:

$\displaystyle \large{\sum_{n=1}^\infty a_1r^{n-1} = \dfrac{a_1}{1-r}}$

Keep in mind that these only work for when |r| < 1 or else it will diverge.

Also, how fortunately, the given summation fits the formula pattern so we do not have to do anything but simply apply the formula in.

$\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = \dfrac{4}{1-0.5}}\\\\\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = \dfrac{4}{0.5}}\\\\\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = 8}$