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Alexxx [7]
3 years ago
9

Please answer to those (a+2)(a+2) (b+7)(b+7) (c-3)(c-3)

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
5 0

Answer:   1. a^2 + 4a + 4

                2. b^2 + 14b + 14

                3. c^2 -6c + 9



vlada-n [284]3 years ago
3 0

Answer: a² + 4a + 4

<u>Step-by-step explanation:</u>

   (a + 2)(a + 2)

= a(a + 2) +2(a + 2)

= a² + 2a + 2a + 4

= a² + 4a + 4

Answer: b² + 14b + 49

<u>Step-by-step explanation:</u>

   (b + 7)(b + 7)

= b(b + 7) +7(b + 7)

= b² + 7b + 7b + 49

= b² + 14b + 49

Answer: c² - 6c + 9

<u>Step-by-step explanation:</u>

   (c - 3)(c - 3)

= c(c - 3) -3(c - 3)

= c² - 3c - 3c + 9

= c² - 6c + 9


SHORTCUT: These are perfect squares. There is a formula for multiplying perfect squares:  (a ± b)² = a² + 2ab + b²

(a + 2)² = a² + 2(a)(2) + (2)²

            = a² + 4a + 4

(b + 7)² = b² + 2(b)(7) + (7)²

           = b² + 14b + 49

(c - 3)² = c² + 2(c)(-3) + (-3)²

           = c² - 6c + 9

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Answer:

1) True 2) False

Step-by-step explanation:

1) Given  \sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}

To verify that the above equality is true or false:

Now find \sum\limits_{k=0}^8\frac{1}{k+3}

Expanding the summation we get

\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3} \sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}

Now find \sum\limits_{i=3}^{11}\frac{1}{i}

Expanding the summation we get

\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}

 Comparing the two series  we get,

\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i} so the given equality is true.

2) Given \sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}

Verify the above equality is true or false

Now find \sum\limits_{k=0}^4\frac{3k+3}{k+6}

Expanding the summation we get

\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}

\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}

now find \sum\limits_{i=1}^3\frac{3i}{i+5}

Expanding the summation we get

\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}

\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}

Comparing the series we get that the given equality is false.

ie, \sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}

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