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valentina_108 [34]

2 years ago

15

An artist wants to frame a square with an area of 400 square inches. She wants to know the length of the wood trim that is neede

d to go around the painting. If x is the length of one side of the painting, what equation can you set up to find the length of a side

1 answer:

gavmur [86]2 years ago

6 0

Multply 400 hudred to the four sides on the lieght

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.. (2.606, 10.837) . . . relative maximum
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2 years ago

For the following amount at the given interest rate compounded continuously, find (a) the future value after 9 years, (b) the

Stella [2.4K]

Answer:

(a) The future value after 9 years is $7142.49.

(b) The effective rate is $r_E=4.759 \:{\%}$ .

(c) The time to reach $13,000 is 21.88 years.

Step-by-step explanation:

The definition of Continuous Compounding is

If a deposit of $P$ dollars is invested at a rate of interest $r$ compounded continuously for $t$ years, the compound amount is

$A=Pe^{rt}$

(a) From the information given

$P=4700$

$r=4.65\%=\frac{4.65}{100} =0.0465$

$t=9 \:years$

Applying the above formula we get that

$A=4700e^{0.0465\cdot 9}\\A=7142.49$

The future value after 9 years is $7142.49.

(b) The effective rate is given by

$r_E=e^r-1$

Therefore,

$r_E=e^{0.0465}-1=0.04759\\r_E=4.759 \:{\%}$

(c) To find the time to reach $13,000, we must solve the equation

$13000=4700e^{0.0465\cdot t}$

$4700e^{0.0465t}=13000\\\\\frac{4700e^{0.0465t}}{4700}=\frac{13000}{4700}\\\\e^{0.0465t}=\frac{130}{47}\\\\\ln \left(e^{0.0465t}\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t\ln \left(e\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t=\ln \left(\frac{130}{47}\right)\\\\t=\frac{\ln \left(\frac{130}{47}\right)}{0.0465}\approx21.88$

3 0

2 years ago

Detmermine the best method to solve the following equation, then solve the equation. (3x-5)^2=-125

liq [111]

For the given equation;

$(3x-5)^2=-125$

We shall begin by expanding the parenthesis on the left side, after which we would combine all terms on and move all of them to the left side, which shall yield a quadratic equation. Then we shall solve.

Let us begin by expanding the parenthesis;

$\begin{gathered} (3x-5)^2\Rightarrow(3x-5)(3x-5) \\ (3x-5)(3x-5)=9x^2-15x-15x+25 \\ (3x-5)^2=9x^2-30x+25 \end{gathered}$

Now that we have expanded the left side of the equation, we would have;

$\begin{gathered} 9x^2-30x+25=-125 \\ \text{Add 125 to both sides and we'll have;} \\ 9x^2-30x+25+125=-125+125 \\ 9x^2-30x+150=0 \end{gathered}$

We shall now solve the resulting quadratic equation using the quadratic formula as follows;

$\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where;} \\ a=9,b=-30,c=150 \\ x=\frac{-(-30)\pm\sqrt[]{(-30)^2-4(9)(150)}}{2(9)} \\ x=\frac{30\pm\sqrt[]{900-5400}}{18} \\ x=\frac{30\pm\sqrt[]{-4500}}{18} \\ x=\frac{30\pm\sqrt[]{-900\times5}}{18} \\ x=\frac{30\pm\sqrt[]{-900}\times\sqrt[]{5}}{18} \\ x=\frac{30\pm30i\sqrt[]{5}}{18} \\ \text{Therefore;} \\ x=\frac{30+30i\sqrt[]{5}}{18},x=\frac{30-30i\sqrt[]{5}}{18} \\ \text{Divide all through by 6, and we'll have;} \\ x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3} \end{gathered}$

ANSWER:

$x=\frac{5+5i\sqrt[]{5}}{3},x=\frac{5-5i\sqrt[]{5}}{3}$

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Step-by-step explanation:

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