How many 5-digit (positive) even numbers can be formed using the digits 4, 6, 7, 2, 8 if digits can be repeated?

1 answer:

4 0

In this problem, since the digits can be repeated, you have to know how many possible digit can be used in every digit of a five – digit number. On the first digit of the number, there are 5 possible digits. The second digit of the number has 5 possible digits as well. The third digit of the number has 5 possible digits. The fourth and the fifth digit will have 5 possible digits as well. Multiplying the possible digits for each digit of a five – digit number:

5 x 5 x 5 x 5 x 5 = 3,125 possible five – digit numbers

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HELP ME Using the following equation, complete the square and find the center and radius of the circle:

IrinaK [193]

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Step-by-step explanation:

Given

the equation is $x^2+y^2+4x-6y-12=0$

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$\Rightarrow (x^2+2\times 2\times x+4-4)+(y^2-2\times3\times y+9-9)-12=0\\\Rightarrow (x^2+2\times 2\times x+4)+(y^2-2\times3\times y+9)-12-4-9=0\\\Rightarrow (x+2)^2+(y-3)^2=25\\\Rightarrow (x+2)^2+(y-3)^2=5^2\\\Rightarrow (x-(-2))^2+(y-3)^2=5^2$

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laila [671]

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3 years ago

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vivado [14]

$2\dfrac{1}{5}+1\dfrac{1}{10}=2+\dfrac{1}{5}+1+\dfrac{1}{10}=(2+1)+\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\\\\=3+\left(\dfrac{1\cdot2}{5\cdot2}+\dfrac{1}{10}\right)=3+\left(\dfrac{2}{10}+\dfrac{1}{10}\right)=3+\dfrac{2+1}{10}=\boxed{3\dfrac{3}{10}}$