. If they set up their camera 322 meters from the base of the mountain and the mountain is 597 meters high, what angle should he
2 answers:
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Megan:
x to the one third power =
<span>x to the one twelfth power = </span>
<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>
<span>
Since </span>
then
Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4
Julie:
x times x to the second times x to the fifth = x * x² * x⁵
<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>![\sqrt[32]{x* x^{2} * x^{5} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B32%5D%7Bx%2A%20x%5E%7B2%7D%20%2A%20x%5E%7B5%7D%20%7D%20)
<span>
Since </span>
Then![\sqrt[32]{x* x^{2} * x^{5} }= \sqrt[32]{ x^{1+2+5} } =\sqrt[32]{ x^{8} }](https://tex.z-dn.net/?f=%5Csqrt%5B32%5D%7Bx%2A%20x%5E%7B2%7D%20%2A%20x%5E%7B5%7D%20%7D%3D%20%5Csqrt%5B32%5D%7B%20x%5E%7B1%2B2%2B5%7D%20%7D%20%3D%5Csqrt%5B32%5D%7B%20x%5E%7B8%7D%20%7D)
Since![\sqrt[n]{x^{m}} = x^{m/n} }](https://tex.z-dn.net/?f=%20%5Csqrt%5Bn%5D%7Bx%5E%7Bm%7D%7D%20%3D%20x%5E%7Bm%2Fn%7D%20%7D%20)
Then![\sqrt[32]{ x^{8} }= x^{8/32} = x^{1/4}](https://tex.z-dn.net/?f=%5Csqrt%5B32%5D%7B%20x%5E%7B8%7D%20%7D%3D%20x%5E%7B8%2F32%7D%20%3D%20x%5E%7B1%2F4%7D%20)
Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.
x to the one third power =
<span>x to the one twelfth power = </span>
<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>
<span>
Since </span>
then
Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4
Julie:
x times x to the second times x to the fifth = x * x² * x⁵
<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>
<span>
Since </span>
Then
Since
Then
Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.
1052 toothpicks can be grouped into 4 groups of third power of 6 (), 5 groups of second power of 6 (
), 1 group of first power of 6 (
) and 2 groups of zeroth power of 6 (
).
The number 1052, written as a base 6 number is 4512
Given: 1052 toothpicks
To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number
First we note that,
This implies that exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.
Now, and
. This implies that
exceeds 1052 and thus there can be at most 4 groups of
.
Then,
So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.
Now, and
. This implies that
exceeds 188 and thus there can be at most 5 groups of
.
Then,
So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.
Now, and
. This implies that
exceeds 8 and thus there can be at most 1 group of
.
Then,
So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.
Now, and
. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.
This concludes the grouping.
Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (), 5 groups of second power of 6 (
), 1 group of first power of 6 (
) and 2 groups of zeroth power of 6 (
).
Then,
So, the number 1052, written as a base 6 number is 4512.
Learn more about change of base of numbers here: