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crimeas [40]
2 years ago
14

You are shopping in a department store. You buy a refrigerator that is originally priced at $1,470. The store is offering a spec

ial deal on this refrigerator and is selling it at 8/10 the original price. How much, in dollars, do you pay for the refrigerator
Mathematics
1 answer:
Mekhanik [1.2K]2 years ago
7 0
Click the link up there
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Graph the function y = 8 –. Which is the best approximation of a point on the function? (4.4, 3.421) (4.6, 4.232) (5.2, 4.025) (
Trava [24]

Answer: (5.9, 3.781)

6 0
3 years ago
How many 2/3 pounds are in a sixth of a pound?
mote1985 [20]
None. ⅔ is greater than 1/6 but ⅔ equals 2 x (1/6).
Basically one half of ⅔ equals 1/6
6 0
3 years ago
I just need help set up
Vilka [71]

Answer:

Step-by-step explanation:

<em>Set up equations (L= Lisa; M = Maria; A = April; J = Jim)</em>

L = 12M

A = 2J

<em>Substitute x for Maria and 6x for Jim</em>

L = 12x

A = 2(6x) = 12x

<em>So, you get:</em>

Maria = x

Jim = 6x

Lisa = 12x

April = 12x

5 0
3 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
Fill in the blanks<br>864+2006=--------+864<br><br>5351 +(574+799)= 574+(5351+_____)​
Mama L [17]

Answer:

2006

799

Step-by-step explanation:

Given the expression:

The right hand side of the sun must be equal to the left hand side ;

Therefore ;

864+2006=--------+864

The sum of 864 and 2006 must be equal to the right hand side sum ; hence

864+2006= 2006 +864

Similarly, the left hand side and right havd side must also be equal here ;

5351 + 574 + 799 = 574 + 5351 + 799

Hence, the missing value is 799

6 0
3 years ago
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