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3 years ago

Help anyone pleaseeee

Mathematics

1 answer:

AlexFokin [52]3 years ago

4 0

Answer:

36 <127

(36,127°)

36 [ cos (127) + i sin (127)]

Step-by-step explanation:

w1 = 2 < 95

w2 = 18 < 32

w1 * w2

We multiply the magnitude and add the angles

w1 * w2 = 2*18 < (95+32)

=36 < 127

36 [ cos (127) + i sin (127)]

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In a group of a hundred and fifty students attending a youth workshop in mombasa, 125 of them are fluent in kiswahili, 135 in en

jek_recluse [69]

Answer:

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

Step-by-step explanation:

Step(i):-

Given total number of students n(T) = 150

Given 125 of them are fluent in Swahili

Let 'S' be the event of fluent in Swahili language

n(S) = 125

The probability that the fluent in Swahili language

$P(S) = \frac{n(S)}{n(T)} = \frac{125}{150} = 0.8333$

Let 'E' be the event of fluent in English language

n(E) = 135

The probability that the fluent in English language

$P(E) = \frac{n(E)}{n(T)} = \frac{135}{150} = 0.9$

n(E∩S) = 95

The probability that the fluent in English and Swahili

$P(SnE) = \frac{n(SnE)}{n(T)} = \frac{95}{150} = 0.633$

Step(ii):-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = P(S) + P(E) - P(S∩E)

= 0.833+0.9-0.633

= 1.1

Final answer:-

The probability that a student chosen at random is fluent in English or Swahili.

P(S∪E) = 1.1

8 0

2 years ago

Find the cross product (7,-4,5) x (6,2,9). is the resulting vector perpendicular to the given vectors ?

Nutka1998 [239]

Answer:

Step-by-step explanation:

On edge

7 0

3 years ago

Can anyone answer this for me?

kramer

Answer:

$\frac{3}{2}$ π (or 270°)

Step-by-step explanation:

I use the trick with the coordinates.

(0, -1)

cos270° = 0 (the x-coordinate)

sin270° = -1 (the y-coordinate)

This trick works depending on the radius.

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3 years ago

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olga_2 [115]

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2 years ago

Find the vertices and foci of the hyperbola with equation quantity x plus 4 squared divided by 9 minus the quantity of y minus 5

My name is Ann [436]

Answer:

Vertices at (-7, 5) and (-1, 5).

Foci at (-9, 5) and (1,5).

Step-by-step explanation:

(x + 4)²/9 - (y - 5)²/16 = 1

The standard form for the equation of a hyperbola with centre (h, k) is

(x - h²)/a² - (y - k)²/b² = 1

Your hyperbola opens left/right, because it is of the form x - y.

Comparing terms, we find that

h = -4, k = 5, a = 3, y = 4

In the general equation, the coordinates of the vertices are at (h ± a, k).

Thus, the vertices of your parabola are at (-7, 5) and (-1, 5).

The foci are at a distance c from the centre, with coordinates (h ± c, k), where c² = a² + b².

c² = 9 + 16 = 25, so c = 5.

The coordinates of the foci are (-9, 5) and (1, 5).

The Figure below shows the graph of the hyperbola with its vertices and foci.

6 0

3 years ago