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alukav5142 [94]
3 years ago
9

Solve the inequality. 14c<8

Mathematics
1 answer:
amid [387]3 years ago
3 0
Divide both sides by 14
c <8/14
c <4/7
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Find all solutions of the equation in the interval [0, 2 pi) 2 cos0-1=0
ddd [48]

Answer:

θ = π/3, 5π/3

Step-by-step explanation:

2 cos θ - 1 = 0

2 cos θ = 1

cos θ = 1/2

θ = π/3, 5π/3

7 0
3 years ago
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If you multiply me by 5 and subtract 6 from me you get 14. what number am I.<br>​
damaskus [11]

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tje answer is 4

Step-by-step explanation:

4x5=20

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27 + 71 = 98 + 1 - ______. What should come in the box to make the above number sentence true?
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2 years ago
HELP MEEEEEEEEEE Package A contains 3 birthday cards and 2 thank-you notes and costs $9.60. Package B contains 8 birthday cards
hodyreva [135]

Answer:

Each birthday card costs $2.2 ⇒ answer c

Step-by-step explanation:

* Lets explain how to solve the problem

- Package A contains 3 birthday cards and 2 thank-you notes

- It costs $9.60

- Package B contains 8 birthday cards and 6 thank-you notes

- It costs $26.60

- x represents the cost of birthday card and y represents the cost of

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* Lets change these information to two equations

∵ x represents the cost of each birthday cards

∵ y represents the cost of each thank-you notes

∵ Bag A contains 3 birthday cards and 2 thank-you notes

∵ Bag A costs $9.60

∴ 3x + 2y = 9.60 ⇒ (1)

∵ Bag B contains 8 birthday cards and 6 thank-you notes

∵ Bag B costs $26.60

∴ 8x + 6y = 26.60 ⇒ (2)

* Lets solve this system of equations to find x and y

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∵ -3(3x) + -3(2y) = -3(9.60)

∴ -9x - 6y = -28.8 ⇒ (3)

- Add equations (2) and (3)

∴ -x = -2.2

- Multiply both sides by -1

∴ x = 2.2

∵ x represents the cost of each birthday cards

∴ The cost of each birthday card is $2.2

* Each birthday card costs $2.2

6 0
3 years ago
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Two fair dice, one blue and one red, are tossed, and the up face on each die is recorded. Define the following events:
ZanzabumX [31]

Answer:

Step-by-step explanation:

Given that two fair dice, one blue and one red, are tossed, and the up face on each die is recorded.

a) P(E) = P(the difference of the numbers is 3 or more}

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P(E) = \frac{12}{36} =\frac{1}{3}

b)P(F)

Favourable events for F = (1,1) (2,2)...(6,6)

P(F) = \frac{6}{36} =\frac{1}{6}

c) P(EF)

There is no common element between E and F

P(EF) =0

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3 years ago
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