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stiv31 [10]
3 years ago
9

PLEASE HELPPPPPPP MEEEEEE

Mathematics
2 answers:
liq [111]3 years ago
5 0

tan (π/6) + sin (5π/3) . cos (- 3π/4)

= tan 30° + sin 300° . cos (- 135°)

= tan 30° + sin (360° - 60°) . cos 135°

= tan 30° + (- sin 60°) . cos (180° - 45°)

= tan 30° + (- sin 60°) . (- cos 45°)

= tan 30° + sin 60° . cos 45°

= (1/3)√3 + (1/2)√3 . (1/2)√2

= (1/3)√3 + (1/4)√6

= (4/12)√3 + (3/12)√6

= (4√3 + 3√6)/12

erik [133]3 years ago
3 0

Answer:

Option 2

Step-by-step explanation:

tan(pi/6) = 1/sqrt(3) = sqrt(3)/3

sin(5pi/3) = -sin(pi/3) = -sqrt(3)/2

cos(-3pi/4) = cos(pi/4) = -1/sqrt(2)

= -sqrt(2)/2

sqrt(3)/3 + -sqrt(3)/2 × -sqrt(2)/2

sqrt(3)/3 + sqrt(6)/4

Lcm: 12

[4sqrt(3) + 3sqrt(6)]/12

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Thanks for all the help
bazaltina [42]

Answer:

b

Step-by-step explanation:

9(y+6)+2y(y+3)/((y+3)(y+6))= optin b

5 0
2 years ago
Solve the inequality. Enter the answer as an inequality that shows the value of the variable; for example f > 7, or 6 < w.
OLEGan [10]
X + 4 > 13 would equal to x > 9
4 0
3 years ago
1- find the first term= ?, 90, 540, 3240
Luda [366]

Answer:

1. 15

2. 8

Step-by-step explanation:

The two sequence are geometric progression GP, because they follow a constant multiple (common ratio)

The nth term of a GP is;

Tn = ar^(n-1)

Where;

a = first term

r = common ratio

For the first sequence;

The common ratio r is

r = T3/T2 = 540/90 = 6

r = 6

T2 = ar^(2-1) = ar

T2 = 90 = ar

Substituting the values of r;

90 = a × 6

a = 90/6

a = 15

First term = 15

2. The sam method applies here.

Common ratio r = T3/T2 = 128/32 = 4

r = 4

T2 = ar^(2-1) = ar

T2 = 32 = ar

Substituting the values of r;

32 = a × 4

a = 32/4

a = 8

First term = 8

4 0
3 years ago
Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,
posledela

Answer:

  • number of multiplies is n!
  • n=10, 3.6 ms
  • n=15, 21.8 min
  • n=20, 77.09 yr
  • n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

  mpy[n] = n·mp[n-1]

  mpy[2] = 2

So, ...

  mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

  10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

  n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

  n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

  n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0
4 years ago
A solid figure without any basses and only one curved surface
Angelina_Jolie [31]
Answer is a sphere (ball shape)
3 0
3 years ago
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