Square root of 120 over the square root of 30. what is the following quotient.

Mathematics

1 answer:

Minchanka [31]2 years ago

3 0

Answer:

21.90890230 or 21.91 (I'm assuming you're saying 120/ √30)

Step-by-step explanation:

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In Exercises 10–14, use the diagram to determine whether you can assume the<br> statement.

lukranit [14]

Answer:

it depends on what he means

Step-by-step explanation:

The friend needs to clarify the meaning of "if three lines intersect each other." If Line A intersects lines B and C, there will be two points of intersection, one at line B and one at line C.

If those lines are all in the same plane, and B and C are not parallel, so that line B intersects line C, then there will be a total of three points of intersection.

If the point of intersection of B and C is also the point where line A intersects them, then there will be only one point of intersection.

So, if the meaning is "if there are three non-parallel lines in the same plane, and each intersects the other two", then the Line Intersection Postulate guarantees there will be 1 or 3 points of intersection.

If the meaning is "if there are three lines not necessarily in the same plane, and one intersects the other two (but those two don't intersect each other)", then there may be 1 or 2 points of intersection (allowing that all lines may intersect at the same point).

7 0

3 years ago

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

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