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3 years ago

Subtract whole number estimate.then find the difference estimate 428,734 -175,842

Mathematics

1 answer:

diamong [38]3 years ago

3 0

You are estimating each into whole numbers. They are already whole numbers, and so just subtract.

428,734 - 175,842 = 252892

252,892 is your answer

hope this helps

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Hello!

Gnesinka [82]

Answer:

A. 2 < x < 10

Step-by-step explanation:

With two congruent sides the third side is greater against a greater angle.

34 < 36, therefore:

3x - 6 < 24
3x < 30
x < 10

And, the angle should have a positive value:

3x - 6 > 0
3x > 6
x > 2

Combined, we get:

2 < x < 10

8 0

3 years ago

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nikitadnepr [17]

Answer:

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

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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

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

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

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