From the first step of the construction, drawing the arcs on both rays establishes that AB is

Mathematics

1 answer:

leonid [27]3 years ago

7 0

Answer: The answer is “Congruent to AC” !! When you are making an arc on both rays, they should measure/be exactly the same, and even more exact is you are using a compass

Step-by-step explanation:

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********PLEASE HELP ****** DVD players are going out of fashion in the modern era. If 400,000 DVD players were sold this year an

abruzzese [7]

So 400,000 DVD sales with a 31% decline means in the 1st year following, there will be a decline of 124,000 DVDs sold. So we need to take 400,000 - 124,000, which equals a starting point for year two is 276,000 DVDs.

So a 31% decline on 276,000 = 85,560. So we need to subtract that from 85,560 from 276,000 which is 190,440. So at the end of the second year, DVD sales were only 190,440. That's also out starting point of the 3rd year.

So 31% of 190,440 equals 59,036.4 for the third year of DVD sales. So we need to subtract 59,036.4 from out third year starting point of 190,440, which equals 131,403.6, but since you can't have parts of DVDs, we'll round the decimal point fraction to a whole number to end up with 131,404 DVDs sold in year three.

So after three years of a 31% yearly decline in DVD sales you end up with DVD sales are 131,404

3 0

3 years ago

Read 2 more answers

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$