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pantera1 [17]
3 years ago
15

How to say 119,000,003 in two other ways?

Mathematics
1 answer:
Sergio039 [100]3 years ago
7 0
You can say it in word form or expanded form
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Find the probability that a randomly
Lerok [7]

Fine the area of the circle:

Area of a circle = pi x r ^2

Area of the circle = 3.14 x 4^2 = 50.24 square cm

Find the area of the triangle:

Area of triangle = 1/2 x base x height

Area =1/2 x 8 x 4 = 16 square cm

Find the area of the white part by subtracting the area of the triangle from the circle:

50.24 - 16 = 34.24 square cm.

The probability of landing in white is the area of white/ area of circle:

34.24/50.24 = 0.682

Multiply by 100 to get percent:

0.682 x 100 = 68.2%

4 0
3 years ago
HELP ME PLEASE I DONT KNOW HOW TO DO THIS
Diano4ka-milaya [45]

Answer:

x = - 6

p = 9

k = 1

Step-by-step explanation:

See attached worksheet.

4 0
2 years ago
4 Tan A/1-Tan^4=Tan2A + Sin2A​
Eva8 [605]

tan(2<em>A</em>) + sin(2<em>A</em>) = sin(2<em>A</em>)/cos(2<em>A</em>) + sin(2<em>A</em>)

• rewrite tan = sin/cos

… = 1/cos(2<em>A</em>) (sin(2<em>A</em>) + sin(2<em>A</em>) cos(2<em>A</em>))

• expand the functions of 2<em>A</em> using the double angle identities

… = 2/(2 cos²(<em>A</em>) - 1) (sin(<em>A</em>) cos(<em>A</em>) + sin(<em>A</em>) cos(<em>A</em>) (cos²(<em>A</em>) - sin²(<em>A</em>)))

• factor out sin(<em>A</em>) cos(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (1 + cos²(<em>A</em>) - sin²(<em>A</em>))

• simplify the last factor using the Pythagorean identity, 1 - sin²(<em>A</em>) = cos²(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (2 cos²(<em>A</em>))

• rearrange terms in the product

… = 2 sin(<em>A</em>) cos(<em>A</em>) (2 cos²(<em>A</em>))/(2 cos²(<em>A</em>) - 1)

• combine the factors of 2 in the numerator to get 4, and divide through the rightmost product by cos²(<em>A</em>)

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - 1/cos²(<em>A</em>))

• rewrite cos = 1/sec, i.e. sec = 1/cos

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - sec²(<em>A</em>))

• divide through again by cos²(<em>A</em>)

… = (4 sin(<em>A</em>)/cos(<em>A</em>)) / (2/cos²(<em>A</em>) - sec²(<em>A</em>)/cos²(<em>A</em>))

• rewrite sin/cos = tan and 1/cos = sec

… = 4 tan(<em>A</em>) / (2 sec²(<em>A</em>) - sec⁴(<em>A</em>))

• factor out sec²(<em>A</em>) in the denominator

… = 4 tan(<em>A</em>) / (sec²(<em>A</em>) (2 - sec²(<em>A</em>)))

• rewrite using the Pythagorean identity, sec²(<em>A</em>) = 1 + tan²(<em>A</em>)

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (2 - (1 + tan²(<em>A</em>))))

• simplify

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (1 - tan²(<em>A</em>)))

• condense the denominator as the difference of squares

… = 4 tan(<em>A</em>) / (1 - tan⁴(<em>A</em>))

(Note that some of these steps are optional or can be done simultaneously)

7 0
3 years ago
Write a polynomial equation with roots 5 and -9i. X^3-?x^2+?X-?=0
jek_recluse [69]

Answer:

x³ - 5x² + 81x - 405 = 0

Step-by-step explanation:

Complex roots occur in conjugate pairs.

Thus given x = - 9i is a root then x = 9i is also a root

The factors are then (x - 5), (x - 9i) and (x + 9i)

The polynomial is the the product of the roots, that is

f(x) = (x - 5)(x - 9i)(x + 9i) ← expand the complex factors

    = (x - 5)(x² - 81i²) → note i² = - 1

    = (x - 5)(x² + 81) ← distribute

    = x³ + 81x - 5x² - 405, thus

x³ - 5x² + 81x - 405 = 0 ← is the polynomial equation

6 0
3 years ago
1) -24, -4, 16, 36, ...<br> Find a23
viktelen [127]

We're given the Arithmetic Progression <em>-24, -4, 16, 36 ...</em> .

We know that a term in an AP is generally represented as:

\bf a_n\ =\ a\ +\ (n\ -\ 1)d

where,

  • a = the first term in the sequence
  • n = the number of the term/number of terms
  • d = difference between two terms

We need to find \sf a_2_3.

From the given progression, we have:

  • a = -24
  • n = 23
  • d = (-24 - (-4) = -20

Using these in the formula,

\sf a_2_3\ =\ a\ +\ (n\ -\ 1)d\\\\\\a_2_3\ =\ -24\ +\ (23\ -\ 1)\ \times\ (-20)\\\\\\a_2_3\ =\ -24\ +\ 22\ \times (-20)\\\\\\a_2_3\ =\ -24\ -\ 440\\\\\\\bf a_2_3\ =\ -464

Therefore, the 23rd term in the AP is -464.

Hope it helps. :)

7 0
3 years ago
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