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1 year ago

Jeremy invests $8, 500 into an account with a 2.4% interest rate that is compounded quarterly. How much money will be in

Mathematics

1 answer:

sertanlavr [38]1 year ago

5 0

Answer:

$9,812.29

Step-by-step explanation:

The amount in Jeremy's account can be computed using the compound interest formula.

<h3>account value</h3>

The formula for the value of an account earning compound interest at annual rate r, compounded n times per year for t years is ...

A = P(1 +r/n)^(nt)

where P is the principal invested.

<h3>formula application</h3>

When P=$8500, r=0.024, n=4, t=6, the formula becomes ...

A = $8500(1 +0.024/4)^(4·6) = $8500(1.006^24) ≈ $9812.29

There will be $9,812.29 in this account after 6 years.

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Factor using the trinomial<br> 10.<br> X2 + 13x + 12

Rudik [331]

Answer:

Factoring the trinomial: $x^2+13x+12$ we get the factors $\mathbf{(x+1)(x+12)}$

Step-by-step explanation:

We need to factor the trinomial: $x^2+13x+12$

We can factor the trinomial by breaking the middle term i.e 13x

Such that adding or subtracting gets the both terms we get 13x and multiplying both terms we get 12x^2

We can break 13x as: 12x and x

Adding them we get 13x and multiplying them we get 12x^2

So,

$x^2+13x+12\\=x^2+12x+x+12\\=x(x+12)+1(x+12)\\=(x+1)(x+12)$

So, Factoring the trinomial: $x^2+13x+12$ we get the factors $\mathbf{(x+1)(x+12)}$

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2 years ago

The area of a circle is 39.62cm2.<br> Find the length of the radius rounded to 2 DP.

Shkiper50 [21]

Using the formula for area of a circle:

Area = pi x radius^2

Replace area with the given value:

39.62 = pi x radius^2

Divide both sides by pi

Radius^2 = 39.62/pi

Radius^2 = 12.62

Solve for radius by taking the square root of both sides

Radius = sqrt(12.62)

Radius = 3.55 cm

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2 years ago

A particular isotope has a half-life of 74 days. If you start with 1 kilogram of this isotope, how much will remain after 150

shtirl [24]

$\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}$

$\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}$