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marshall27 [118]
2 years ago
6

Find a polynomial of lowest degree that has the indicated zeroes: 1+i, 1-i, 3

Mathematics
1 answer:
KiRa [710]2 years ago
8 0

Answer:

-3i + (6 x 9)

Step-by-step explanation:

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Use each integer only once. You must have each row, column, and diagonal adding up to the magic sum.
Grace [21]
For question number 3
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5, -5, -6, 8

For question number 4:
-4, 1, -10, 3,
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5, -8, -1, -6,
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8 0
2 years ago
Read 2 more answers
Please help. How do you find cosine, sine, cosecant and secant with this triangle? ​
Veseljchak [2.6K]

Hi there! You have to remember these 6 basic Trigonometric Ratios which are:

  • sine (sin) = opposite/hypotenuse
  • cosine (cos) = adjacent/hypotenuse
  • tangent (tan) = opposite/adjacent
  • cosecant (cosec/csc) = hypotenuse/opposite
  • secant (sec) = hypotenuse/adjacent
  • cotangent (cot) = adjacent/opposite
  • cosecant is the reciprocal of sine
  • secant is the reciprocal of cosine
  • cotangent is the reciprocal of tangent

Back to the question. Assuming that the question asks you to find the cosine, sine, cosecant and secant of angle theta.

What we have now are:

  • Trigonometric Ratio
  • Adjacent = 12
  • Opposite = 10

Looks like we are missing the hypotenuse. Do you remember the Pythagorean Theorem? Recall it!

  • a²+b² = c²

Define that c-term is the hypotenuse. a-term and b-term can be defined as adjacent or opposite

Since we know the value of adjacent and opposite, we can use the formula to find the hypotenuse.

  • 10²+12² = c²
  • 100+144 = c²
  • 244 = c²

Thus, the hypotenuse is:

\large \boxed{c = 2 \sqrt{61} }

Now that we know all lengths of the triangle, we can find the ratio. Recall Trigonometric Ratio above! Therefore, the answers are:

  • cosine (cosθ) = adjacent/hypotenuse = 12/(2√61) = 6/√61 = <u>(6√61) / 61</u>
  • sine (sinθ) = opposite/hypotenuse = 10/(2√61) = 5/√61 = <u>(5√61) / 61</u>
  • cosecant (cscθ) is reciprocal of sine (sinθ). Hence, cscθ = (2√61/10) = <u>√61/5</u>
  • secant (secθ) is reciprocal of cosine (cosθ). Hence, secθ = (2√61)/12 = <u>√</u><u>61</u><u>/</u><u>6</u>

Questions can be asked through comment.

Furthermore, we can use Trigonometric Identity to find the hypotenuse instead of Pythagorean Theorem.

Hope this helps, and Happy Learning! :)

5 0
3 years ago
. Let X and Y be random variables of possible percent returns (0%, 10%,
bazaltina [42]

(a) The marginal distribution of <em>X</em> is

Pr(<em>X</em> = <em>x</em>) = ∑ Pr(<em>X</em> = <em>x</em>, <em>Y</em> = <em>y</em>)

… = 0.0625 + 0.0625 + 0.0625 + 0.0625

… = 0.25

That is, the first equality follows from the law of total probability, with the sum taken over <em>y</em> from {0, 5, 10, 15}. Each probability Pr(<em>X</em> = <em>x</em>, <em>Y</em> = <em>y</em>) is given in the table to be 0.0625.

Similarly, the marginal distribution of <em>Y</em> is

Pr(<em>Y</em> = <em>y</em>) = 0.25

(b) Yes, they're independent because

Pr(<em>X</em> = <em>x</em>, <em>Y</em> = <em>y</em>) = 0.0625,

and

Pr(<em>X</em> = <em>x</em>) Pr(<em>Y</em> = <em>y</em>) = 0.25 • 0.25 = 0.0625.

(c) The mean of <em>X</em> is

E[<em>X</em>] = ∑ <em>x</em> Pr(<em>X</em> = <em>x</em>)

… = 0.25 ∑ <em>x</em>

<em>… </em>= 0.25 (0 + 5 + 10 + 15)

… = 7.5

and you would find the same mean for <em>Y</em>,

E[<em>Y</em>] = 7.5

The variance of <em>X</em> is

V[<em>X</em>] = E[<em>X</em>^2] - E[<em>X</em>]^2

… = (∑ <em>x</em>^2 Pr(<em>X</em> = <em>x</em>)) - 7.5^2

… = 0.25 (∑ <em>x</em>^2) - 56.25

… = 0.25 (0^2 + 5^2 + 10^2 + 15^2) - 56.25

… = 31.25

and similarly,

V[<em>Y</em>] = 31.25

(each sum is taken with <em>x</em> and <em>y</em> from {0, 5, 10, 15})

7 0
3 years ago
7 1/4 boxes weigh total of 66 7/8 what does one box weigh
Ray Of Light [21]

Answer:

68.875 / 7.25 = 9 1/2 pounds

Step-by-step explanation:

6 0
3 years ago
An online music store offers 5 downloads for $6.25. Another online music store offers 12 downloads for $17.40. Which store offer
babymother [125]
6.25
explanation: if you divide 6.25 by 5 you get 1.25, if you divide 17.40 by 12 it’s 1.45. Therefore you pay more.
6 0
2 years ago
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