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enyata [817]
2 years ago
7

Is this correct please help will mark brainliest

Mathematics
1 answer:
zlopas [31]2 years ago
4 0

Answer:

yes i believe it is correct

Step-by-step explanation:

You might be interested in
Evaluate the limit
wel

We are given with a limit and we need to find it's value so let's start !!!!

{\quad \qquad \blacktriangleright \blacktriangleright \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}

But , before starting , let's recall an identity which is the <em>main key</em> to answer this question

  • {\boxed{\bf{a^{2}-b^{2}=(a+b)(a-b)}}}

Consider The limit ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}

Now as directly putting the limit will lead to <em>indeterminate form 0/0.</em> So , <em>Rationalizing</em> the <em>numerator</em> i.e multiplying both numerator and denominator by the <em>conjugate of numerator </em>

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}\times \dfrac{\sqrt{x}+\sqrt{3\sqrt{x}-2}}{\sqrt{x}+\sqrt{3\sqrt{x}-2}}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-\sqrt{3\sqrt{x}-2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}{(x^{2}-4^{2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Using the above algebraic identity ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x})^{2}-(\sqrt{3\sqrt{x}-2})^{2}}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-(3\sqrt{x}-2)}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}+2}{\{(\sqrt{x})^{2}-2^{2}\}(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , here we <em>need</em> to <em>eliminate (√x-2)</em> from the denominator somehow , or the limit will again be <em>indeterminate </em>,so if you think <em>carefully</em> as <em>I thought</em> after <em>seeing the question</em> i.e what if we <em>add 4 and subtract 4</em> in <em>numerator</em> ? So let's try !

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2+4-4}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(x-4)+2+4-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , using the same above identity ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+6-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+3(2-\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , take minus sign common in <em>numerator</em> from 2nd term , so that we can <em>take (√x-2) common</em> from both terms

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)-3(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , take<em> (√x-2) common</em> in numerator ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)\{(\sqrt{x}+2)-3\}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Cancelling the <em>radical</em> that makes our <em>limit again and again</em> <em>indeterminate</em> ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\cancel{(\sqrt{x}-2)}\{(\sqrt{x}+2)-3\}}{\cancel{(\sqrt{x}-2)}(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}+2-3)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-1)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , <em>putting the limit ;</em>

{:\implies \quad \sf \dfrac{\sqrt{4}-1}{(\sqrt{4}+2)(4+4)(\sqrt{4}+\sqrt{3\sqrt{4}-2})}}

{:\implies \quad \sf \dfrac{2-1}{(2+2)(4+4)(2+\sqrt{3\times 2-2})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{6-2})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{4})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+2)}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(4)}}

{:\implies \quad \sf \dfrac{1}{128}}

{:\implies \quad \bf \therefore \underline{\underline{\displaystyle \bf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}=\dfrac{1}{128}}}}

3 0
2 years ago
Read 2 more answers
Divide the following polynomials using the long division model (4x ^ 4 - 5x ^ 3 + 2x ^ 2 - x + 5) + (x ^ 2 + x + 1)
densk [106]

Answer:

The ramainder is equal -86

Step-by-step explanation:

Since the polynomial degree Q(x) is 1, the remainder of the division must be a number. Therefore we only need to calculate the value of polynomial for x = -2

5 0
3 years ago
You have 800 square feet of the room reserved for tables
Sav [38]
I hope this helps for question a=16
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8 0
3 years ago
The perimeter of a rectangle is twice the
Shkiper50 [21]

Answer:

L=16 W=7

Step-by-step explanation:

P=2(L+W)

L=2W+2

46=2(2W+2)+2W

(((7×2)+2)=16×2=32

(16-2)/2=7×2=14

32+14=46 or 16+16+7+7=46

3 0
3 years ago
Find the distance between M(6, -7) and a line that contains<br> C(-6, -2) and H(-1, 10).
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Answer

Step-by-step explanation:

i really need help with it

4 0
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