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ella [17]
3 years ago
15

Help plz I really need the help​

Mathematics
1 answer:
Colt1911 [192]3 years ago
3 0

Answer:

the answer might be either one of the four

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What values complete each statement?
aliya0001 [1]

For this case we have:

By properties of the radicals \sqrt {a} = a ^{\frac {1} {2}}

So:

(\sqrt {7}) ^ 2 = (7 ^ {\frac {1} {2}}) ^ 2.

Now, for power properties we have:

(b ^ {\frac {c} {d}}) ^ e = b ^ {\frac {c * e} {d}}

Thus, (7 ^ {\frac {1} {2}}) ^ 2 = 7 ^ {\frac {2} {2}} = 7

So:

7 ^ {\frac {1} {2}} = \sqrt {7}in its radical form

Answer:

(\sqrt {7}) ^ 2 = (7 ^ {\frac {1} {2}}) ^ 2= 7 ^ {\frac {2} {2}} = 7 in its simplest form.

7 ^ {\frac {1} {2}} = \sqrt {7}in its radical form


4 0
2 years ago
Which number is rational?<br> √2<br> 10<br> 016
erica [24]
16 I think is a rational number
6 0
3 years ago
Use place value pattern to complete the table
svp [43]
Do you have a picture
4 0
3 years ago
The table shows the price of two cereal brands and the number of ounces per box. Which is the better price per ounce?
Romashka-Z-Leto [24]

Answer:

A) Cereal brand A

B) There is no table.

Step-by-step explanation:

A) divide each cereal price by the amount of ounces. Brand A is $0.13888 (rounds to $0.14). B is $0.14583 (rounds to $0.15) So, since cereal band A is less cost per ounce, its cheaper

B) This question cannot be answered, since there is no table.

5 0
2 years ago
<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%
xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }

Let a = 1 and b the cosine product, and write them as

\dfrac{a - b}{x^2}

with

b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}

Now we use the identity

a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)

to rationalize the numerator. This gives

\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}

For the remaining limit, use the Taylor expansion for cos(x) :

\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)

where \mathcal{O}(x^4) essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2

\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)

\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0
2 years ago
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