To evaluate the series we proceed as follows:
2. -5,-15,-25,-35,-45
The common difference is (-15--5)=-10
thus the series is an arithmetic sequence hence the explicit formula will be:
an=a+(n-1)d
thus the formula for the series is:
an=-5+(n-1)(-10)
an=-10n+5
4. 0.5, 0.25,0,...,-0.75
The common difference =0.25-0.5=-0.25
The sequence is an arithmetic sequence:
the explicit formula will be:
an=0.5-0.25(n-1)
an=-0.25n+0.75
thus
n=4
a4=-0.25*4+0.75=-0.25
when n=5
a5=-0.25*5+0.75=-0.5
when n=6
a6=-0.25*6+0.75=-0.75
the missing values are -0.25 and -0.5
6] 4.5,5.6,6.7, ....,11.1
common difference=5.6-4.5=1.1
the sequence is an arithmetic sequence
The explicit formula is:
an=4.5+1.1(n-1)
an=1.1n+3.4
when n=4
an=1.1*4+3.4
a4=7.8
when n=5
a5=1.1*5+3.4
a5=8.9
a6=1.1*6+3.4
a6=10
the missing values are: 7.8, 8.9, 10
let's change some the 0.1 to say 1/10, just the fraction version of it.

![\bf \cfrac{-10x-1}{-10x^3-x^2}\implies \cfrac{-10\left( \frac{1}{10} \right)-1}{-10\left( \frac{1}{10} \right)^3-\left( \frac{1}{10} \right)^2}\implies \cfrac{-1-1}{-\frac{1}{100}-\frac{1}{100}}\implies \cfrac{-2}{\frac{-2}{100}} \\\\\\ \cfrac{~~\begin{matrix} -2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{1}\cdot \cfrac{100}{~~\begin{matrix} -2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies 100](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B-10x-1%7D%7B-10x%5E3-x%5E2%7D%5Cimplies%20%5Ccfrac%7B-10%5Cleft%28%20%5Cfrac%7B1%7D%7B10%7D%20%5Cright%29-1%7D%7B-10%5Cleft%28%20%5Cfrac%7B1%7D%7B10%7D%20%5Cright%29%5E3-%5Cleft%28%20%5Cfrac%7B1%7D%7B10%7D%20%5Cright%29%5E2%7D%5Cimplies%20%5Ccfrac%7B-1-1%7D%7B-%5Cfrac%7B1%7D%7B100%7D-%5Cfrac%7B1%7D%7B100%7D%7D%5Cimplies%20%5Ccfrac%7B-2%7D%7B%5Cfrac%7B-2%7D%7B100%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B~~%5Cbegin%7Bmatrix%7D%20-2%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%7B1%7D%5Ccdot%20%5Ccfrac%7B100%7D%7B~~%5Cbegin%7Bmatrix%7D%20-2%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%5Cimplies%20100)
when checking an absolute value expression, we do the one-sided limits, since an absolute value expression is in effect a piecewise function with ± versions, so for the limit from the left we check the negative version.
Answer:

General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
<u>Algebra I</u>
- Terms/Coefficients
- Factoring
<u>Algebra II</u>
- Exponential Rule [Powering]:

- Solving exponential equations
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>
<em />
<em />
<em />
<u>Step 2: Solve for </u><em><u>x</u></em>
- Rewrite:

- Set:

- Factor:

- [Division Property of Equality] Divide 3 on both sides:

- [Subtraction Property of Equality] Subtract 3x on both sides:

- [Subtraction Property of Equality] Subtract 6 on both sides:

- [Division Property of Equality] Divide -1 on both sides:

Step-by-step explanation:
The formula for the area of a square is:
A
=
s
2
Where:
A
is the area of the square.
s
is the length of the side of a square.
Substituting and solving for
s
gives:
225
cm
2
=
s
2
We can take the square root of each side of the equation giving:
√
225
cm
2
=
√
s
2
15
cm
=
s
s
=
15
cm
The formula for the perimeter of a square is:
p
=
4
s
Where:
p
is the perimeter of the square.
s
is the length of the side of a square.
Substituting for
s
from the solution for the previous formula and calculating
p
gives:
p
=
4
×
15
cm
p=60cm
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<h3>I'll teach you how to solve (81x^-20 y^-16)^-3/4</h3>
---------------------------------------------------------------
(81x^-20 y^-16)^-3/4
Apply exponents rule:
(81x^-20 y^-16)^-3/4=
1 / (81x^-20 y^-16)^-3/4
Simplify (81x^-20 y^-16)^-3/4:
1/27/x^15 y^12
Apply the fraction rule:
x^15 y^12 / 27
Your Answer Is x^15 y^12 / 27
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