1 poir

Mathematics

1 answer:

poizon [28]3 years ago

8 0

9514 1404 393

Answer:

$1790.99

Step-by-step explanation:

Given:

$1625 is invested at an annual rate of 1.95% compounded quarterly for 5 years

Find:

the ending balance

Solution:

The compound interest formula applies.

FV = P(1 +r/n)^(nt) . . . Principal P at rate r for t years, compounded n per year

FV = $1625(1 +0.0195/4)^(4·5) = $1625(1.004875^20) ≈ $1790.99

The account ending balance would be $1790.99.

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postnew [5]

Answer:

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Step-by-step explanation:

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Katyanochek1 [597]

Write tan in terms of sin and cos.

$\displaystyle \lim_{t\to0}\frac{\tan(6t)}{\sin(2t)} = \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)}$

Recall that

$\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1$

Rewrite and expand the given limand as the product

$\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)$

Then using the known limit above, it follows that

$\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}$