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2011 1 .
2025-05-09 11:16 WEIG .
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IGFPX BLG.
- IgIgGGraphitic IgIg.
8.13 BLG vs AL IG .
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T1 3:0 IG 3:0IG T1BO5T1T1.
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IG HLE Doinb.

2011 1 .
2025-05-09 11:16 WEIG .
iG NIP2.
IGFPX BLG.
8.13 BLG vs AL IG .
igblgig ig25ig
T1 3:0 IG 3:0IG T1BO5T1T1.
AL .
IG HLE Doinb.
Solving \(x^2 + 50x - 6000 = 0\) using the quadratic formula: \(x = \frac{-50 \pm \sqrt{50^2 + 4 \times 6000}}{2}\).
\(x = \frac{-50 \pm \sqrt{2500 + 24000}}{2} = \frac{-50 \pm \sqrt{26500}}{2}\).
\(x = \frac{-50 \pm 162.63}{2}\). Taking the positive root: \(x = 56.315\) meters.
Length is \(56.315 + 50 = 106.315\) meters.
#### Width: 56.315 m, Length: 106.315 m
A sum of money is invested at 5% annual interest, compounded annually. If the investment grows to $10,500 in 3 years, what was the initial amount?
Let the initial amount be \(P\). The formula is \(P(1 + r)^n = 10500\).
Substituting \(r = 0.05\) and \(n = 3\): \(P(1.05)^3 = 10500\).
\(P \times 1.157625 = 10500\).
Solving for \(P\), \(P = \frac{10500}{1.157625} \approx 9070.58\).