Finance Unit
1.5 Order in Calculations/Evaluating Formulas
Simplify each expression. Show each step one at a time.
- \(7-24 \div 3 \times 4+6\)
\(7-8 \times 4+6\)
\(7-32+6\)
\(-25+6\)
\(-19\)
- \(20(3+7)\)
\(20(10)\)
\(200\)
- \(12+4^{2}\)
\(12+16\)
\(28\)
- \(6\left(5-3\right)^{2}\)
\(6\left(2\right)^2\)
\(6(4)\)
\(24\)
- \(200 \times(1+0.04 / 12)^{5 \times 12}\)
\(200 \cdot\left(1+\frac{1}{300}\right)^{60}\)
\(200 \cdot\left(\frac{301}{300}\right)^{60}\)
\(200 \cdot 1.2210\)
\(244.20\)
- \(50 \times \frac{\left[(1+0.05 / 12)^{12 \times 10}-1\right]}{(0.05 / 12)}\)
\(7764.11\)
- \(\frac{180000 \times(0.065 / 12)}{\left[1-(1+0.065 / 12)^{(-12 \times 15)}\right]}\)
\(1567.99\)
- \(\pi(8)^{2}\)
\(64 \pi\)
\(201.0619\)
- \(\frac{4}{3} \pi(7.5)^3\)
\(\frac{4}{3} \pi \left(421.875\right)\)
\(562.5 \pi\)
\(1767.1459\)
- \(\frac{1}{2} \times 15(3.2+10.8)\)
\(\frac{1}{2} \times 15(14)\)
\(7.5 \cdot 14\)
\(105\)
- Evaluate \(I=\operatorname{Prt} \text { when } P=2000, r=0.045, \text { and } t=\frac{11}{12}\)
\(I=2000 \cdot 0.045 \cdot \frac{11}{12}\)
\(I=\$82.50\)
- Find the simple interest earned on $3000 at an annual interest rate of 3.75% for 4 years. Use the simple interest formula \(I=Prt\) where \(P\) is the principle invested, \(r\) is the annual interest rate in decimal form, and \(t\) is the number of years.
\(I=3000 \cdot 0.0375 \cdot 4\)
\(I=\$450\)
- Evaluate \(A=P\left(1+\frac{r}{n}\right)^{n t} \text { when } P=2000, r=0.045, n=12, \text { and } t=\frac{11}{12}\)
\(A=2000\left(1+\frac{.045}{12}\right)^{12 \cdot \frac{11}{12}}\)
\(\$2084.06\)
- Find the accumulated amount when $3000 is invested for 4 years at a 3.75% interest rate compounded monthly. Use the compound interest formula \(A=P\left(1+\frac{r}{n}\right)^{n t}\) where \(P\) is the principle invested, \(r\) is the annual interest rate in decimal form, \(t\) is the number of years, and \(n\) is the number of compounding periods in a year.
\(A=3000\left(1+\frac{.0375}{12}\right)^{12 \cdot 4}\)
\(A=\$3484.69\)
- Evaluate \(A=P M T \times \frac{\left[\left(1+\frac{R}{n}\right)^{n t}-1\right]}{\left(\frac{R}{n}\right)} \text { when } P M T=\$ 250, R=0.02, t=20, \text { and } n=12\)
\(A=\frac{250\left[\left(1+{\displaystyle\frac{.02}{12}}\right)^{12\ast20}-1\right]}{\left({\displaystyle\frac{.02}{12}}\right)}\)
\(A=\$73699.21\)
- Find the accumulated amount when $1000 is invested every month for 3.5 years at a 3% interest rate compounded monthly. Use the savings plan formula \(A=P M T \times \frac{\left[\left(1+\frac{R}{n}\right)^{n t}-1\right]}{\left(\frac{R}{n}\right)}\) where \(PMT\) is the monthly payment, \(R\) is the annual percentage rate, \(t\) is the number of years, and \(n\) is the number of compounding periods per year.
\(A=\frac{1000\left[\left(1+{\displaystyle\frac{.03}{12}}\right)^{12\ast3.5}-1\right]}{\left({\displaystyle\frac{.03}{12}}\right)}\)
\(A=\$44226.03\)
- Evaluate \(a^{2}+b^{2}=c^{2} \text { when } a=5 \text { and } b=12\)
\(5^{2}+12^{2}=c^{2}\)
\(c^{2}=25+144\)
\(c^{2}=169\)
\(\sqrt{c^2}=\sqrt{169}\)
\(c=13\)
- Find the length of the hypotenuse of a right triangle when the legs are 14 cm and 48 cm. Use the Pythagorean Theorem \(a^{2}+b^{2}=c^{2}\).
\(14^{2}+48^{2}=c^{2}\)
\(c^{2}=2500\)
\(c=50\) cm
- Evaluate \(x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\) when \(a = 3, b = -6,\) and \(c = 2\). Find both solutions to the nearest hundredth.
\(x=\frac{6 \pm \sqrt{(-6)^{2}-4(3)(2)}}{2 \cdot 3}\)
\(x=\frac{6 \pm \sqrt{12}}{6}\)
\(x \approx 1.58,0.42\)
- Find the solutions to the equation \(-2 x^{2}+12 x+10=0\). The equation is in standard form \(a x^{2}+b x+c=0\). Use the quadratic formula \(x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\) to solve for x. Find both solutions to the nearest hundredth.
\(x=\frac{-12 \pm \sqrt{12^{2}-4(-2)(10)}}{2 \cdot-2}\)
\(x=\frac{-12 \pm \sqrt{224}}{-4}\)
\(x \approx-0.74,6.74\)