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wariber [46]
3 years ago
7

An introductory APR is the interest that applies to _____.

Mathematics
2 answers:
mafiozo [28]3 years ago
8 0

Answer: The answer is any combination of purchases, balance transfers and cash advances.

Step-by-step explanation:  An introductory APR rate is defined as follows:

An introductory APR rate is defined as a rate given by credit card issuers to new customers for a particular period. APR rate is very low or most often 0% that applies to any combination of purchases, balance transfers and cash advances depending on the details provided by the by the issuer.

Thus, the answer is  any combination of purchases, balance transfers, and cash advances.

olganol [36]3 years ago
5 0
<span>An introductory APR is the interest that applies to </span>new applicants of a credit card as an incentive to apply for the card; offered by the credit card company.

After the introductory period is over, the APR will go up.
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Circumference of the circle
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37 5/7

Step-by-step explanation:

use formular it will be 2pie -22/7 ×radius

since we have been given only radius it will be

2×22/7×6=37 5/7

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What is the correct letter? (Giving brainliest)
vampirchik [111]

Answer:

d is the answer

Step-by-step explanation:

perimeter of rect = 2(7+3) = 20

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2 years ago
Damian is going to invest in an account paying an interest rate of 5.2% compounded
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650

Step-by-step explanation:

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3 years ago
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Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)
leonid [27]
Let P=(x,y,z) be an arbitrary point on the surface. The distance between P and the given point (7,11,0) is given by the function

d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}

Note that f(x) and f(x)^2 attain their extrema, if they have any, at the same values of x. This allows us to consider the modified distance function,

d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2

So now you're minimizing d^*(x,y,z) subject to the constraint z^2=xy. This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)

which has partial derivatives

\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}

Setting all four equation equal to 0, you find from the third equation that either z=0 or \lambda=-1. In the first case, you arrive at a possible critical point of (0,0,0). In the second, plugging \lambda=-1 into the first two equations gives

\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10

and plugging these into the last equation gives

z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5

So you have three potential points to check: (0,0,0), (2,10,2\sqrt5), and (2,10,-2\sqrt5). Evaluating either distance function (I use d^*), you find that

d^*(0,0,0)=170
d^*(2,10,2\sqrt5)=46
d^*(2,10,-2\sqrt5)=46

So the two points on the surface z^2=xy closest to the point (7,11,0) are (2,10,\pm2\sqrt5).
5 0
3 years ago
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