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Ainat [17]
2 years ago
8

X2-x-6=0, then x is​

Mathematics
2 answers:
Leni [432]2 years ago
8 0

Answer:

A is the answers for the question

Step-by-step explanation:

please mark me as brainlest

musickatia [10]2 years ago
8 0

\underline{\underline{\large\bf{Solution:-}}}\\

\longrightarrowGiven is a quadratic equation which can be solved by splitting the middle term.

\leadstoFor this we have to look at factors of the constant term . Since here constant term term is- 6 so splitted middle term should have product equal to -6

\begin{gathered}\\\implies\quad \sf x^2-x-6=0 \\\end{gathered}

\begin{gathered}\\\implies\quad \sf x^2-3x+2x-6=0 \\\end{gathered}

\begin{gathered}\\\implies\quad \sf x(x-3)+2(x-3)=0 \\\end{gathered}

\begin{gathered}\\\implies\quad \sf (x-3)(x+2) =0 \\\end{gathered}

\begin{gathered}\\\implies\quad \sf (x-3) =0 \quad or \quad (x+2) =0\\\end{gathered}

\begin{gathered}\\\implies\quad \sf x=3 \quad or \quad x= -2\\\\\end{gathered}

\quad\therefore\: \sf x  \:is \: \:3  \:or  -2

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olga nikolaevna [1]
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If y=\ln x, then we have -1\le \ln x\le1\implies \dfrac1e\le x\le e as the domain of \cos^{-1}(\ln x).

We know that if f and g are continuous functions, then so is the composite function f\circ g.

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7 0
3 years ago
Use the following image to assist in your answer.<br> a =<br> - 6<br> - 9<br> - 4
vfiekz [6]

Step-by-step explanation:

Pythagoras' theorem for the smallest one :

c^2 = 4^2 + 6^2

c^2 = 16 + 36

c^{2} = 52

Pythagoras' theorem for the middle one :

b^{2} = 6^{2} + a^{2}

Pythagoras' theorem for the biggest one :

(4+a)^2 = c^2 + b^2

16 + 8a + a^2 = 52 + b^2

Using the formula before (for b^2) it becomes :

16 + 8a + a^2 = 52 + (6^2 + a^2)

16 + 8a + a^2 = 52 + 6^2 + a^2

16 + 8a = 52 + 6^2

16 + 8a = 52 + 36

16+8a = 88

8a = 88-16

8a = 72

a = \frac{72}{8}

a = 9

Verifying :

b^2 = 6^2 + a^2

b^2 = 36 + 81

b^2 = 117

b^{2} = 117

The biggest one :

(4+a)^2 = c^2 + b^2

(4+9)^2 = 52 + 117

13^2 = 169

True

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Answer:

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I inserted an image of the equation.

Step-by-step explanation:

Hope this helps!

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