Answer:
<h3>C. They are both perfect squares and perfect cubes.</h3>
Step-by-step explanation:
Perfect squares are numbers that their square root can be found easily without any remainder.
Given the following patterns;
1*1 = 1 and 1*1*1 = 1
It can be seen that 1 is 1 perfect square since 1*1 = 1² = 1
Also 1 is perfect cube since 1*1*1 = 1³ = 1 (cube of the value gives 1)
Similarly for the expression;
8*8 = 64
8² = 64 (since the square of 8 gives 64, then 64 is known to be a perfect square)
Also 4*4*4 = 64
i.e 4³ = 64 (This shows that the cube root of 64 is 4 making it a perfect cube since we can get a whole number for the cube root of 64)
The same is applicable for other expressions 729 = 27 × 27, and 9 × 9 × 9, 4,096 = 64 × 64, and 16 × 16 × 16
This values are easily expressed as a constant multiple of a number showing that they are both perfect squares and perfect cubes.
Answer:
85,999,999.999 999 909
Step-by-step explanation:
The expression represents the difference of a relatively large number and one that is relatively small. That difference is approximately the value of the large number. The exact value requires 17 digits for its proper expression. Most calculators and spreadsheets cannot display this many digits.
<h3>Standard form</h3>
The numbers in standard form are ...
86,000,000 = 8.6×10^7
0.000000091 = 9.1×10^-8
<h3>Difference</h3>
Their difference is ...
86,000,000 -0.000000091 = 85,999,999.999 999 909
In scientific notation, this is ...
8.599 999 999 999 990 9×10^7
Hey!
Let's walk through this together.
Possible Answers: {5, 7, 9, 11, 13}
Let's start with 5.
5-4<8
1<8
So, 5 is CORRECT.
Next, 7.
7-4<8
3<8
So, 7 is CORRECT.
Next, 9.
9-4<8
5<8
So, 9 is CORRECT.
Next, 11.
11-4<8
7<8
So, 11 is CORRECT.
Last, 13.
13-4<8
11<8
So, 13 is INCORRECT.
Your answers are: {5,7,9,11}
So D.
Hope this helps!
<h2><em>
~~~PicklePoppers~~~</em></h2>
Answer:
We accept the null hypothesis and conclude that voltage for these networks is 232 V.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 232 V
Sample mean, 
= 231.5 V
Sample size, n = 66
Sample standard deviation, s = 2.19 V
Alpha, α = 0.05
First, we design the null and the alternate hypothesis
We use Two-tailed t test to perform this hypothesis.
Formula:
Putting all the values, we have
Now,
Since,
We accept the null hypothesis and conclude that voltage for these networks is 232 V.
All area : 8x*12x=96x^2
Area for field : 3x*6x=18x^2
Area of land wich will be left for the bleachers,restrooms,and other parts of the stadium: 96x^2-18x^2=<span>78x^2
</span>So the answer is A.<span>78x^2</span>
