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3 years ago

How can we get rid of i in the denominator?

Mathematics

1 answer:

Sloan [31]3 years ago

8 0

Answer:

When you have an imaginary number in the denominator multiply the numerator and the denominator by the conjugate of the denominator.

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Darnell team scored 40 points at the game Donald score 15% of the teams points how many points the Darnell score

andriy [413]

Donald scores 6 points

3 0

3 years ago

If f(x) = 4x - 12, what is f(2)? O A. -4 O'B. 4 O c. -20 D. 8 Helpppp plzzz

photoshop1234 [79]

Answer:

-4

Step-by-step explanation:

f(x) = 4x - 12

Let x=2

f(2) = 4*2 - 12

= 8-12

= -4

8 0

3 years ago

John find 1/5 of the bar chocolate he bought was already eaten by his brother. He eat 3/4 of what was remaining. Find the portio

larisa86 [58]

Answer:

1/5 is left.

Step-by-step explanation:

Portion remaining after the bar was eaten by his brother = 1 - 1/5 = 4/5.

Portion remaining after John ate 3/4 of this = 4/5 - 3/4* 4/5 = 4/5 - 3/5

= 1/5.

3 0

3 years ago

A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 2

Shalnov [3]

Answer:

The expected winnings for a person buying 1 ticket is -0.2.

Step-by-step explanation:

Given : A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 25 cents each, find the expected winnings for a person buying 1 ticket.

To find : What are the expected winnings?

Solution :

There are one first prize, 2 second prize and 20 third prizes.

Probability of getting first prize is $\frac{1}{20000}$

Probability of getting second prize is $\frac{2}{20000}$

Probability of getting third prize is $\frac{20}{20000}$

A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each.

So, The value of prizes is

$\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10$

If 20000 tickets are sold at 25 cents each i.e. $0.25.

Remaining tickets = 20000-1-2-20=19977

Probability of getting remaining tickets is $\frac{19977}{20000}$

The expected value is

$E=\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10-\frac{19977}{20000}\times 0.25$

$E=\frac{1000+600+200-4994.25}{20000}$

$E=\frac{-3194.25}{20000}$

$E=-0.159$

Therefore, The expected winnings for a person buying 1 ticket is -0.2.

3 0

3 years ago

Find the value of x

fgiga [73]

$\qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star$

x = 35°

$\textsf{ \underline{\underline{Steps to solve the problem} }:}$

$\qquad❖ \: \sf \:4x + 1 + x + 4 = 180$

[ by linear pair ]

$\qquad❖ \: \sf \:5x + 5 = 180$

$\qquad❖ \: \sf \:5(x + 1) = 180$

$\qquad❖ \: \sf \:x + 1 = 180 \div 5$

$\qquad❖ \: \sf \:x + 1 = 36$

$\qquad❖ \: \sf \:x = 36 - 1$

$\qquad❖ \: \sf \:x = 35$

$\qquad \large \sf {Conclusion} :$

x = 35°

7 0

2 years ago