What should I do to get the correct answer?
1 answer:
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There are two ways to work this out: normal variables or using "imaginary" numbers.
Normal variables:
![(7+2i)(3-i)\\(7*3)+[7*(-i)]+(3*2i)+[2i*(-i)]\\21-7i+6i-2i^{2}\\\\21-i-2i^{2}](https://tex.z-dn.net/?f=%20%287%2B2i%29%283-i%29%5C%5C%287%2A3%29%2B%5B7%2A%28-i%29%5D%2B%283%2A2i%29%2B%5B2i%2A%28-i%29%5D%5C%5C21-7i%2B6i-2i%5E%7B2%7D%5C%5C%5C%5C21-i-2i%5E%7B2%7D)
Imaginary numbers:
Using the result from earlier:
Now since
, then the expression becomes:
Normal variables:
Imaginary numbers:
Using the result from earlier:
Now since
9 longer sticks were used.
This is how i get it.
1. change m to cm, 1m =100cm
2. add both together, divide by 16 =8.125. Which means 1 of the stick should more than 8, and 1 of the other stick should be less than 8
3. substitute the equation
4.(55 x 7)+(75 x 9) =1060
This is how i get it.
1. change m to cm, 1m =100cm
2. add both together, divide by 16 =8.125. Which means 1 of the stick should more than 8, and 1 of the other stick should be less than 8
3. substitute the equation
4.(55 x 7)+(75 x 9) =1060
Answer:
there are only 4 whole numbers whose squares and cubes have the same number of digits.
Explanations:
let 0, 1, 2 and 4∈W (where W is a whole number), then
,
,
,
,
,
,
,
.
You can see from the above that only four whole numbers are there whose squares and cubes have the same number of digits