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3 years ago

6.5x-10=4.5x+6 can someone give a step by step answer plz

Mathematics

2 answers:

scoundrel [369]3 years ago

8 0

Answer:

$\huge{\bold{\boxed{\sf{x = 8 }}}}$

Step-by-step explanation:

$6.5x - 10 = 4.5x + 6$

Move 4.5x to left hand side and change its sign.

Similarly, move 10 to right hand side and change its sign.

⇒ $6.5x - 4.5x = 6 + 10$

Subtract 4.5x from 6.5x

⇒ $2x = 6 + 10$

Add the numbers: 6 and 10

⇒ $2x = 16$

Divide both sides by 2

⇒ $\frac{2x}{2} = \frac{16}{2}$

Calculate

⇒ $\boxed{\sf{ x = 8 }}$

Hope I helped!

Best regards!

~ $\sf{TheAnimeGirl}$

Trava [24]3 years ago

7 0

Collect like terms
6.5x-4.5x = 6+10
2x = 6+10
Divide both sides
2x = 16
X= 8

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3 years ago

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madreJ [45]

The expression as given doesn't make much sense. I think you're trying to describe an infinitely nested radical. We can express this recursively by

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Then you want to know the value of

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To show the limit exists and that $a_n$ converges to some limit, we can try showing that the sequence is bounded and monotonic.

Boundedness: It's true that $a_1=\sqrt{42}\le\sqrt{49}=7$ . Suppose $a_k\le 7$ . Then $a_{k+1}=\sqrt{42+a_k}\le\sqrt{42+7}=7$ . So by induction, $a_n$ is bounded above by 7 for all $n$ .

Monontonicity: We have $a_1=\sqrt{42}$ and $a_2=\sqrt{42+\sqrt{42}}$ . It should be quite clear that $a_2>a_1$ . Suppose $a_k>a_{k-1}$ . Then $a_{k+1}=\sqrt{42+a_k}>\sqrt{42+a_{k-1}}=a_k$ . So by induction, $a_n$ is monotonically increasing.