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Tamiku [17]
3 years ago
6

How many different outcomes are possible if you can choose between a bus or a train, full fare or concession and you can travel

in the morning or the afternoon?
Mathematics
1 answer:
Novosadov [1.4K]3 years ago
7 0
3 pairs of choices give rise to 2^3 = 8 possibilities:
.. bus, full, morning
.. bus, full, afternoon
.. bus, concession, morning
.. bus, concession, afternoon
.. train, full, morning
.. train, full, afternoon
.. train, concession, morning
.. train, concession, afternoon
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Factorise and solve : x^3-x^2-4x+4
Alika [10]

(x - 1)(x - 2)(x + 2)

note that the sum of the coefficients 1 - 1 - 4 + 4 = 0

thus x = 1 is a root and (x - 1 ) is a factor

dividing x³ - x² - 4x + 4 by (x - 1)

x³ - x² - 4x + 4 = (x - 1)(x² - 4 ) (note (x² - 4 ) is a difference of squares )

x³ - x² - 4x + 4 = (x - 1)(x - 2)(x + 2)

(x - 1)(x - 2)(x + 2 ) =0

x - 1 = 0 ⇒ x = 1

x - 2 = 0 ⇒ x = 2

x + 2 = 0 ⇒ x = - 2

solutions are x = 1 or x = ± 2



6 0
3 years ago
Simplify each expression.<br><br> 5x+ 3/4 +2x− 1/2
suter [353]

Answer:

7x+1/4

Step-by-step explanation:

5x+3/4+2x+-1/2

Combine Like Terms:

5x+3/4+2x+-1/2

(5x+2x)+(3/4+-1/2)

7x+1/4

Hope this helps

7 0
3 years ago
What is the solution to x-4=17
Serhud [2]
17+4 = 21 the answer is 21
3 0
3 years ago
Read 2 more answers
show the sample space of a toss of two dice as a list denote the event of a sum of 3 or 6 or 9 on the list
Ratling [72]
The sample space is:
(1, 1); (1, 2) - sum of 3; (1, 3); (1, 4); (1, 5) - sum of 6; (1, 6);
(2, 1) - sum of 3; (2, 2); (2, 3); (2, 4) - sum of 6; (2, 5); (2, 6);
(3, 1); (3, 2); (3, 3) - sum of 6; (3, 4); (3, 5); (3, 6) - sum of 9;
(4, 1); (4, 2) - sum of 6; (4, 3); (4, 4); (4, 5) - sum of 9; (4, 6);
(5, 1) - sum of 6; (5, 2); (5, 3); (5, 4) - sum of 9; (5, 5); (5, 6);
(6, 1): (6, 2); (6, 3) - sum of 9; (6, 4); (6, 5); (6, 6)
4 0
3 years ago
Need help with this asap, please
Lilit [14]

Answer:

<h3>             Perimeter = 3√5 + 6 + √73 + √58 + 5√2</h3>

Step-by-step explanation:

AB=\sqrt{(6-0)^2+(5-8)^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt5\\\\BC=|-1-5|=|-6|=6\\\\CD=\sqrt{(-2-6)^2+(-4-(-1))^2}=\sqrt{64+9}=\sqrt{73}\\\\DE=\sqrt{(-5-(-2))^2+(3-(-4))^2}=\sqrt{9+49}=\sqrt{58}\\\\EA=\sqrt{(0-(-5))^2+(8-3)^2}=\sqrt{25+25}=\sqrt{25\cdot2}=5\sqrt2

6 0
3 years ago
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