Answer:
2%
Step-by-step explanation:
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:
f(2n)-f(n)=log2
b.lg(lg2+lgn)-lglgn
c. f(2n)/f(n)=2
d.2nlg2+nlgn
e.f(2n)/(n)=4
f.f(2n)/f(n)=8
g. f(2n)/f(n)=2
Step-by-step explanation:
What is the effect in the time required to solve a prob- lem when you double the size of the input from n to 2n, assuming that the number of milliseconds the algorithm uses to solve the problem with input size n is each of these function? [Express your answer in the simplest form pos- sible, either as a ratio or a difference. Your answer may be a function of n or a constant.]
from a
f(n)=logn
f(2n)=lg(2n)
f(2n)-f(n)=log2n-logn
lo(2*n)=lg2+lgn-lgn
f(2n)-f(n)=lg2+lgn-lgn
f(2n)-f(n)=log2
2.f(n)=lglgn
F(2n)=lglg2n
f(2n)-f(n)=lglg2n-lglgn
lg2n=lg2+lgn
lg(lg2+lgn)-lglgn
3.f(n)=100n
f(2n)=100(2n)
f(2n)/f(n)=200n/100n
f(2n)/f(n)=2
the time will double
4.f(n)=nlgn
f(2n)=2nlg2n
f(2n)-f(n)=2nlg2n-nlgn
f(2n)-f(n)=2n(lg2+lgn)-nlgn
2nLg2+2nlgn-nlgn
2nlg2+nlgn
5.we shall look for the ratio
f(n)=n^2
f(2n)=2n^2
f(2n)/(n)=2n^2/n^2
f(2n)/(n)=4n^2/n^2
f(2n)/(n)=4
the time will be times 4 the initial tiote tat ratio are used because it will be easier to calculate and compare
6.n^3
f(n)=n^3
f(2n)=(2n)^3
f(2n)/f(n)=(2n)^3/n^3
f(2n)/f(n)=8
the ratio will be times 8 the initial
7.2n
f(n)=2n
f(2n)=2(2n)
f(2n)/f(n)=2(2n)/2n
f(2n)/f(n)=2
Selection B is appropriate.
_____
It takes 12 times as long to fill the whole tank as it does to fill 1/12 of the tank.
.. 12 * (1/3 hour) = (1/3)*12 hour
First mug holds the most
<em><u>Solution:</u></em>
Given that,
You are choosing between two mugs
<em><u>The volume of cylinder is given as:</u></em>

Where,
r is the radius and h is the height
<em><u>One has a base that is 5.5 inches in diameter and a height of 3 inches</u></em>

Therefore,

Also, h = 3 inches
<em><u>Thus volume of cylinder is given as:</u></em>

Thus first mug holds 71.24 cubic inches
<em><u>The other has a base of 4.5 inches in diameter and a height of 4 inches</u></em>

h = 4 inches
Therefore,

Thus the second mug holds 63.585 cubic inches
On comparing, volume of both mugs,
Volume of first mug > volume of second mug
First mug holds the most