Thousands grou
the thousands group. 913,256 256 being the ones group 913 being the thousands group. the name of the period with 913, is the thousands group p
Answer:
d.
Step-by-step explanation:
To convert a root to a fraction in the exponent, remember this rule:
![\sqrt[n]{a^{m}}=a^{\frac{m}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5E%7Bm%7D%7D%3Da%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D)
The index becomes the denominator in the fraction. (The index is the little number in front of the root, "n".) The original exponent remains in the numerator.
In this question, the index is 4.
The index is applied to every base in the equation under the root. The bases are 16, 'x' and 'y'.
![\sqrt[4]{16x^{15}y^{17}} = (\sqrt[4]{16})(\sqrt[4]{x^{15}})(\sqrt[4]{y^{17}}) = (2)(x^{\frac{15}{4}}})(y^{\frac{17}{4}}) = 2x^{\frac{15}{4}}}y^{\frac{17}{4}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B16x%5E%7B15%7Dy%5E%7B17%7D%7D%20%3D%20%28%5Csqrt%5B4%5D%7B16%7D%29%28%5Csqrt%5B4%5D%7Bx%5E%7B15%7D%7D%29%28%5Csqrt%5B4%5D%7By%5E%7B17%7D%7D%29%20%3D%20%282%29%28x%5E%7B%5Cfrac%7B15%7D%7B4%7D%7D%7D%29%28y%5E%7B%5Cfrac%7B17%7D%7B4%7D%7D%29%20%3D%202x%5E%7B%5Cfrac%7B15%7D%7B4%7D%7D%7Dy%5E%7B%5Cfrac%7B17%7D%7B4%7D%7D)
To find the quad root of 16, input this into your calculator. Since 2⁴ = 16,
= 2.
For the "x" and "y" bases, use the rule for converting roots to exponent fractions. The index, 4, becomes the denominator in each fraction.

Answer:
$ 20
Step-by-step explanation:
Given that,
No. of notes Joe has = 4
The total value of the notes = $ 75
From the given notes, 4 notes that can make the sum of 75 would be
$ 50 note + $ 10 + $ 10 note + $ 5
= 50 * 1 + 10 * 1 + 10 * 1 + 5 * 1
= $ 75
Thus, the note that would not be included in the total value would $20 note as if we add it either the sum would be greater or lesser with four notes in combination.
Since GL and HI are parallel to each other, then in the translation this condition is going to be kept. So that, we can write in these gaps that AB and FE are parallel to each other.
Step-by-step explanation:
