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3 years ago

PLEASE HELP ME!!!!!!

Mathematics

1 answer:

Sonja [21]3 years ago

6 0

Answer:

$\huge \red{ c_3 = - 1}$

Step-by-step explanation:

$c_1 = 1 \\ \\ c_n= - 2c_{n-1}+5 \\ \\ c_2 = - 2c_{2-1}+5 \\ \\ c_2 = - 2c_{1}+5 \\ \\ c_2 = - 2(1)+5 \\ \\ c_2 = - 2+5 \\ \\ \huge \purple{c_2 =3} \\ \\ c_3 = - 2c_{3-1}+5 \\ \\ c_3 = - 2c_{2}+5 \\ \\ c_3 = - 2(3)+5 \\ \\ c_3= - 6+5 \\ \\ \huge \red{ c_3 = - 1}$

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2 years ago

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The missing value in the logarithm are as follows:

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<h3>How to solve logarithm?</h3>

Using logarithm rule,

logₐ b - logₐ c = logₐ (b / c)

logₐ b + logₐ c = logₐ (b × c)

Therefore,

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learn more on logarithm here: brainly.com/question/24515212

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2 years ago

What number must you add to complete the square?<br> x^2 + 2x = -1

Andru [333]

$\large\begin{array}{l} \textsf{You must add 1 to complete the square.}\\\\ \textsf{There is a very common special product: (square of a sum)}\\\\ \mathsf{(a+b)^2=a^2+2ab+b^2}\\\\\\ \textsf{Take the given equation:}\\\\ \mathsf{x^2+2x=-1\qquad(i)}\\\\\\ \textsf{The first term is a square (x squared)}\\\\ \textsf{The second term is a product of 2, x and 1.} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{We need to find a proper value for b, so that if we add it to}\\\textsf{the left side of the equation, we get a perfect square.}\\\\ \textsf{So,}\\\\ \begin{array}{cccccccccc} \mathsf{a^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{a}&\!\!\!\cdot\!\!\!&\mathsf{b}&\!\!\!+\!\!\!&\mathsf{b^2}&=\mathsf{(a+b)^2}\\\\ \mathsf{\downarrow}&&&\!\!\!\!\!&\downarrow&\!\!\!\!\!&\downarrow&&\downarrow\\\\ \mathsf{x^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{x}&\!\!\!\cdot\!\!\!&\mathsf{1}&\!\!\!+\!\!\!&\mathsf{1^2}&=\mathsf{(x+1)^2} \end{array}\\\\\\ \textsf{Just by comparing the expressions above, we can conclude that}\\\textsf{b must be equal to 1, so we get a perfect square:}\\\\ \mathsf{(x+1)^2=x^2+2x+1} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{Back to (i), adding 1 to both sides:}\\\\ \mathsf{x^2+2x+1=-1+1}\\\\ \boxed{\begin{array}{c} \mathsf{(x+1)^2=0} \end{array}} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{If you want to solve this equation for x, you will find}\\\\ \mathsf{x+1=0}\\\\ \mathsf{x=-1}\quad\longleftarrow\quad\textsf{(solution)} \end{array}$

If you're having problems understanding the answer, try seeing it through your browser: brainly.com/question/2112248

$\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best regards! :-)} \end{array}$

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3 years ago

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