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3 years ago

Lol please help :))))).

Mathematics

1 answer:

Pani-rosa [81]3 years ago

7 0

C and D are irrational.

Generally, any square root that doesn't reduce to a rational number is irrational. Some examples:

• √2 is irrational

• √4 = 2 is rational

• √(1/2) = 1/√2 = √2/2 is irrational

• √(1/9) = 1/√9 = 1/3 is rational

Numbers with non-terminating decimal expansions are rational if there is a repeating pattern after the decimal. The pattern in 4.737737773… is pretty obvious (one 7, one 3, two 7s, one 3, three 7s, one 3, etc) but it does not repeat, the number of 7s just keeps going up. Compare this to 4.737373… or 4.773773773… - these two numbers are rational in contrast.

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What is 10.26 with 8% tax

sleet_krkn [62]

Answer:

$11.08 to the nearest penny

Step-by-step explanation:

So I do it like

10.26 is 100 percent

We’re adding 8 percent so technically multiplying by 108

SO 10.26*108 is 11.08 to the nearest penny

It so much easier, easy 1 step

8 0

3 years ago

Read 2 more answers

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 main key 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 indeterminate form 0/0. So , Rationalizing the numerator i.e multiplying both numerator and denominator by the conjugate of numerator 

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

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

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 need to eliminate (√x-2) from the denominator somehow , or the limit will again be indeterminate ,so if you think carefully as I thought after seeing the question i.e what if we add 4 and subtract 4 in numerator ? 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 numerator from 2nd term , so that we can take (√x-2) common 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 (√x-2) common 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 radical that makes our limit again and again indeterminate ;

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

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

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

Now , putting the limit ;

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

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

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

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

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

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

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

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

3 0

2 years ago

Read 2 more answers

Jerry multiplied 3.04 x 7.091 and got the product 2.155664. Is his answer reasonable? Why or why not?

Readme [11.4K]

First we multiply the numbers ignoring the decimal and then place the decimal in the position

3.04 *7.091 = 2155664

Now the first number has two digits after the decimal and the second number has three digits after the decimal.

Therefore in the final result we have to move the decimal (2+3) =5 place from the left,

So the answer would be 21.55664

So, the correct answer is 21.55664

Jerry calculated the answer correctly , but made an error in the placement of the decimal point. The answer is not reasonable because 2.155664 and 21.55664 mean two completely different answer far apart from each other

6 0

3 years ago

PLEASE HELP ASAP!!! What is the value of x? Enter your answer in the box x =

poizon [28]

Answer:Titus is going to invest $500. Bank A offers a simple interest rate of 4%, while Bank B offers an interest rate of 3% compounded annually. In the long run, after many years, which bank account will grow the largest?

Step-by-step explanation:

8 0

3 years ago

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Find the area of the figure to the nearest thousandth

riadik2000 [5.3K]

Answer:

Total Area = 236.715 sq ft

Step-by-step explanation:

this is a composite figure made up of a circle and a rectangle; find the area of each and then add them together

A(circle) = πr²

= π(7.5²)

= 176.715

A(rectangle) = 15(4) which is 60

Total Area = 236.715 sq ft

7 0

3 years ago