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White raven [17]
4 years ago
6

Simpilify the expresssion 2+3i/4+3i

Mathematics
1 answer:
Eddi Din [679]4 years ago
3 0
Multiply the numerator and denominator by 4+5i
=[(2+3i)(4+5i)]/[(4-5i)(4+5i)] 
=[(8-15)+(12+10)i] / [16+25] <span>= [ -7 + 22i ] / [41] 
= (-7/41) + (22/41)i </span>
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3 years ago
George has $6.40 in nickels and dimes. If he has
VLD [36.1K]

Answer:

George has 64 nickel and 32 dimes.

Step-by-step explanation:

Normally, we have:

One nickel = 5 cents

One dime = 10 cents

One dollar = 100 cents

Therefore, total number of cents that George has can be calculated as follows:

Total number of cents = $6.40 * 100 = 640 cents

Based on the above, we have:

640 cents = 640 / 5 = 128 nickel

640 cents = 640 / 10 = 64 dimes

Therefore, we have:

128 nickel = 64 dimes

Divide through by 2 in order to share 640 cents equally, we have:

128 nickel / 2 = 64 dimes / 2 => 64 nickel = 32 dimes

Since 64 minus 32 is equal to 32, it therefore implies that George has 64 nickel and 32 dimes.

3 0
3 years ago
How to solve x(x-3)=0
damaskus [11]

Step-by-step explanation:

x(x-3)=0

x=0 or (x-3) =0

x= 0 or x =3

7 0
3 years ago
Earth travels around the sun at a speed of about 30 kilometers per second. Convert this speed to kilometers per minute.
Verizon [17]

Answer:

1800

Step-by-step explanation:

there are 60 seconds in a minute

so, 30 * 60 = 1800

8 0
3 years ago
Lotteries are an important income source for various governments around the world. However, the availability of lotteries and ot
SashulF [63]

Answer: 0.0473

Step-by-step explanation:

Given : The proportion of gambling addicts : p=0.30

Let x be the binomial variable that represents the number of persons are gambling addicts.

with parameter p=0.30 , n= 10

Using Binomial formula ,

P(x)=^nC_xp^x(1-p)^{n-x}

The required probability = P(x>5)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\=^{10}C_6(0.3)^6(0.7)^{4}+^{10}C_7(0.3)^7(0.7)^{3}+^{10}C_8(0.3)^8(0.7)^{2}+^{10}C_9(0.3)^9(0.7)^{1}+^{10}C_{10}(0.3)^{10}(0.7)^{0}\\\\=\dfrac{10!}{4!6!}(0.3)^6(0.7)^{4}+\dfrac{10!}{3!7!}(0.3)^7(0.7)^{3}+\dfrac{10!}{2!8!}(0.3)^8(0.7)^{2}+\dfrac{10!}{9!1!}(0.3)^9(0.7)^{1}+(1)(0.3)^{10}\\\\=0.0473489874\approx0.0473

Hence, the required probability = 0.0473

5 0
3 years ago
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