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Usimov [2.4K]
3 years ago
11

How do I simplify this?​

Mathematics
1 answer:
vladimir1956 [14]3 years ago
8 0

Answer:

Domain: all real numbers except 0 and -3

Vertical asymptotes: x = 0; x = -3

Step-by-step explanation:

The domain is the set of numbers that can be substituted for x.

The expression has x in the denominator. Since division by zero is undefined, x cannot have any value that would cause a zero in the denominator.  To find which values of x cause a zero in the denominator, we set the denominator equal to zero and solve the equation for x.

2x^2 + 6x = 0

Factor 2x.

2x(x + 3) = 0

Divide both sides by 2.

x(x + 3) = 0

x = 0 or x + 3 = 0

x = 0 or x = -3

Since the values x = 0 and x = -3 cause the denominator to equal zero, these two values of x must be excluded from the domain.

Domain: all real numbers except 0 and -3

The vertical asymptotes happen at the x values excluded from the domain.

Vertical asymptotes: x = 0; x = -3

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The cost of 6 movie tickets is $57.00. What is the cost of 8 movie tickets?
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Step-by-step explanation:

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What is the sign of the product (−4)(2)(−3)(6)?
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Answer:

The sign would be positive.

The answer would be 144.

Step-by-step explanation:

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2 years ago
Find the 2th term of the expansion of (a-b)^4.​
vladimir1956 [14]

The second term of the expansion is -4a^3b.

Solution:

Given expression:

(a-b)^4

To find the second term of the expansion.

(a-b)^4

Using Binomial theorem,

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here, a = a and b = –b

$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}

Substitute i = 0, we get

$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4

Substitute i = 1, we get

$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b

Substitute i = 2, we get

$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}

Substitute i = 3, we get

$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}

Substitute i = 4, we get

$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}

Therefore,

$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}

=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}

Hence the second term of the expansion is -4a^3b.

3 0
3 years ago
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