H E R E W E G O
1=1.61
x10 x10
10 = 16.1
16.1 kilometers he drove
L = 3 + 2w
Find the width
Area = 54
l × w = 54
(3 + 2w) × w = 54
3w + 2w^2 = 54
2w^2 + 3w - 54 = 0
(2w - 9)(w + 6) = 0
w = 9/2 or w = -6 (width shouldn't be negative)
w = 9/2
w = 4.5 m
Find the length
l = 3 + 2w
l = 3 + 2(4.5)
l = 3 + 9
l = 12 m
The width is 4.5 m, the length is 12 m
<h3>
Answer: 1</h3>
where x is nonzero
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Explanation:
We'll use two rules here
- (a^b)^c = a^(b*c) ... multiply exponents
- a^b*a^c = a^(b+c) ... add exponents
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The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.
Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)
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After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2), (b^2-c^2) and (c^2-a^2)
Add up those exponents (using rule 2 above) and we get
(a^2-b^2)+(b^2-c^2)+(c^2-a^2)
a^2-b^2+b^2-c^2+c^2-a^2
(a^2-a^2) + (-b^2+b^2) + (-c^2+c^2)
0a^2 + 0b^2 + 0c^2
0+0+0
0
All three exponents add to 0. As long as x is nonzero, then x^0 = 1
Solutions
In Matrix we use initially based on systems of linear equations.The matrix method is similar to the method of Elimination as but is a lot cleaner than the elimination method.Solving systems of equations by Matrix Method involves expressing the system of equations in form of a matrix and then reducing that matrix into what is known as Row Echelon Form.<span>
Calculations
</span>⇒ <span>Rewrite the linear equations above as a matrix
</span>
⇒ Apply to Row₂ : Row₂ - 2 <span>Row₁
</span>
⇒ <span>Simplify rows
</span>
Note: The matrix is now in echelon form.
<span>The steps below are for back substitution.
</span>
⇒ Apply to Row₁<span> : Row</span>₁<span> - </span>5 Row₂
⇒ <span>Simplify rows
</span>
⇒ <span>Therefore,
</span>

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