heureka

How much of a radioactive substance must be presented to decay to 60 grams in 11 years if the half-life of the substance is 3.9 years?

\(\begin{array}{rcll} A(t)&=&A_0\cdot \left(\dfrac12 \right)^\dfrac{t}{t_{1/2}},\ \text{where} \\ \\ A(t) &-& \text{the amount left after t years;} \\ A_0 &-& \text{the initial quantity of the substance that will undergo decay;} \\ t_{1/2} &-& \text{the half-life of the decaying quantity.} \\ \end{array} \)

\(\begin{array}{|rcll|} \hline A(t) &=& 60g \\ t_{1/2} &=& 3.9 y \\ t &=& 11y \\\\ \hline A(t)&=& A_0\cdot \left(\dfrac12 \right)^\dfrac{t}{t_{1/2}} \\ 60g &=& A_0\cdot \left(\dfrac12 \right)^\dfrac{11y}{3.9 y} \\ 60g &=& A_0\cdot \left(\dfrac12 \right)^\dfrac{11 }{3.9 } \\ 60g &=& A_0\cdot \left(\dfrac12 \right)^{2.82051282051} \\ 60g &=& A_0\cdot 0.5^{2.82051282051} \\ 60g &=& A_0\cdot 0.14156015773 \\\\ A_0 &=& \dfrac{60g}{0.14156015773} \\\\ \mathbf{A_0} & \mathbf{=} & \mathbf{423.848072521g} \\ \hline \end{array}\)

b. Between 420 and 440 grams

yes, sorry

but the calculation is okay.

(6cos²24°-6sin²24°)tan48°

\(\begin{array}{|rcll|} \hline &&\mathbf{ (6\cos^224^\circ-6\sin^224^\circ)\tan48^\circ } \\\\ &=& 6\cdot (\cos^224^\circ-\sin^224^\circ)\tan(48^\circ) \quad | \quad \cos(48^\circ) = \cos^2(24^\circ) - \sin^2(24^\circ) \\\\ &=& 6\cdot \cos(48^\circ)\cdot \tan(48^\circ) \quad | \quad \tan(48^\circ) = \dfrac{\sin(48^\circ)}{\cos(48^\circ)} \\\\ &=& 6\cdot \cos(48^\circ)\cdot \dfrac{\sin(48^\circ)}{\cos(48^\circ)} \\\\ &=& 6\cdot \dfrac{\sin(48^\circ)\cos(48^\circ)}{\cos(48^\circ)} \\\\ &\mathbf{=}& \mathbf{6\cdot \sin(48^\circ)} \\ \hline \end{array} \)

Evaluate the sum

\(\large{\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots}\)

\(\begin{array}{|rcll|} \hline && \dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots \\\\ &=& \dfrac{6}{(3-1)(3+1)}+\dfrac{6}{(5-1)(5+1)}+\dfrac{6}{(7-1)(7+1)}+\dfrac{6}{(9-1)(9+1)}+\cdots \\\\ &=& \dfrac{6}{2\cdot 4}+\dfrac{6}{4\cdot 6}+\dfrac{6}{6\cdot 8}+\dfrac{6}{8\cdot 10}+\cdots \\\\ &=& 6\left( \dfrac{1}{2\cdot 4}+\dfrac{1}{4\cdot 6}+\dfrac{1}{6\cdot 8}+\dfrac{1}{8\cdot 10}+\cdots \right) \\\\ &=& 6\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{2k(2k+2)} \\\\ &=& 6\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{4k(k+1)} \\\\ &=& \dfrac{6}{4}\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{k(k+1)} \\\\ &=& \dfrac{3}{2}\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{k(k+1)} \quad | \quad \boxed{\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1} } \\\\ &=& \dfrac{3}{2}\cdot \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k} - \dfrac{1}{k+1} \right) \\\\ &=& \dfrac{3}{2}\cdot \left[\sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k} \right) - \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k+1} \right) \right] \\\\ &=& \dfrac{3}{2}\cdot \left[1+\sum \limits_{k=2}^{\infty} \left(\dfrac{1}{k} \right) - \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k+1} \right) \right] \\\\ &=& \dfrac{3}{2}\cdot \left[1+\sum \limits_{k=2}^{\infty} \left(\dfrac{1}{k} \right) - \sum \limits_{k=2}^{\infty} \left(\dfrac{1}{k} \right) \right] \\\\ &=& \dfrac{3}{2}\cdot (1+0) \\\\ &=& \dfrac{3}{2} \\ \hline \end{array}\)

Thank you, CPhill !

We have a parabola with equation 

\(y=-2{x^2}+8x+10\),

and a hyperbola with equation 

\(xy=16\)

We have to solve  \(\mathbf{ x^3-4x^2-5x +8=0}\)

1.

We attempt x = 1

\(\begin{array}{|rcll|} \hline \mathbf{ x^3-4x^2-5x+8} &\mathbf{=}& \mathbf{0} \quad | \quad x=1 \\\\ 1^3-4\cdot 1^2-5\cdot 1 +8 &\overset{?}{=} & 0 \\ 1 -4 -5 +8 &\overset{?}{=} & 0 \\ 9-9 &\overset{?}{=} & 0 \\ 0 & = & 0\ \checkmark \\ \hline \end{array}\)

We have found the first root, it is \( x = 1 \quad (1.\text{ quadrant })\)

\(\begin{array}{|rcll|} \hline x^3-4x^2-5x+8 = ( ~?~ )(x-1) \\ (x^3-4x^2-5x+8) : (x-1) = ( ~?~ ) \\ \hline \end{array} \)

2.

Polynomial long division:

\(\begin{array}{|rcll|} \hline x^3-4x^2-5x+8 = ( x^2 - 3x-8 )(x-1) &=& 0 \\\\ x^2 - 3x -8 &=& 0 \\ x &=& \dfrac{3 \pm \sqrt{3^2-4(-8) } } {2} \\ x &=& \dfrac{3 \pm \sqrt{9+32} } {2} \\ x &=& \dfrac{3 \pm \sqrt{41} } {2} \\\\ x_2 &=& \dfrac{3 + \sqrt{41} } {2} \\ x_2 &=& 4.70156211872 \quad | \quad 1.\text{ quadrant } \\\\ x_3 &=& \dfrac{3 - \sqrt{41} } {2} \\ \mathbf{ x_3 } &\mathbf{=}& \mathbf{-1.70156211872} \quad | \quad 3.\text{ quadrant } \\\\ y_3 &=& \dfrac{16}{x_3} \\ y_3 &=& \dfrac{16}{-1.70156211872} \\ \mathbf{ y_3 } &\mathbf{=}& \mathbf{-9.40312423743} \\ \hline \end{array}\)

The point of intersection in the 3th quadrant is \(\mathbf{(-1.70156211872, -9.40312423743)}\)

Hi jm,

\(\begin{array}{|rcll|} \hline y &=& -2x^2+8x+10 \quad & \quad \cdot x \\ yx &=& -2x^3+8x^2+10x \quad & \quad xy = 16 \\ 16 &=& -2x^3+8x^2+10x \quad & \quad :2 \\ 8 &=& -x^3+4x^2+5x \quad & \quad \cdot (-1) \\ -8 &=& x^3-4x^2-5x \\ \mathbf{ x^3-4x^2-5x} &\mathbf{=}& \mathbf{-8} \\ \hline \end{array}\)

I get  \(x^3-4x^2-5x= -8\), why do you get  \(x^3-4{x^2}-5x=16\) ?

3.

In the diagram below,  \(AM=BM=CM ={\color{red}\mathbf{r} } \) and  \(\angle BMC+\angle A = 201^\circ\)

Find  \(\angle B\)  in degrees.

\(\text{$\triangle ABC$ is a right-angled triangle, so $\angle A = 90^\circ - \angle B$ } \\ \text{$\angle CMA = 2\angle B$ }\\ \text{$\angle BMC = 180^\circ - \angle CMA$ }\)

\(\begin{array}{|rcll|} \hline \angle BMC+\angle A &=& 201^\circ \quad | \quad BMC = 180^\circ - \angle CMA \\ 180^\circ - \angle CMA+\angle A &=& 201^\circ \quad | \quad CMA = 2\angle B \\ 180^\circ - 2\angle B+\angle A &=& 201^\circ \quad | \quad A = 90^\circ - \angle B \\ 180^\circ - 2\angle B+90^\circ - \angle B &=& 201^\circ \\ 270^\circ - 3\angle B &=& 201^\circ \\ 3\angle B &=& 270^\circ- 201^\circ \\ 3\angle B &=& 69^\circ \\ \mathbf{\angle B} & \mathbf{=} & \mathbf{23^\circ} \\ \hline \end{array}\)

\(\angle B\)  in degrees is \(\mathbf{23^\circ}\)

i have to find is the point of inetrsection in the 4th quadrant...

There is no intersection in the 4th quadrant, while \(y=\dfrac{4}{x}\) is only in the 1st and 3th quadrant!

The line y = 12 - 2x is a tangent to two curves.

Each curve has an equation of the form y= k + 6 + kx - x^2, where k is a constant.

Find the two values of k.

The Point at the tangent:

\(\begin{array}{|rcll|} \hline y = 12-2x &=& k+6+kx-x^2 \\ 12-2x &=& k+6+kx-x^2 \\ x^2-(2+k)x+6-k &=& 0 \\\\ x &=& \dfrac{2+k\pm \sqrt{(2+k)^2-4(6-k) } } {2} \\ x &=& \dfrac{2+k\pm \sqrt{4+4k+k^2-24+4k } } {2} \\ \mathbf{x} & \mathbf{=} & \mathbf{\dfrac{2+k\pm \sqrt{k^2+8k-20 } } {2}} \\ \hline \end{array} \)

The slope at the tangent:

\(\begin{array}{|rcll|} \hline y &=& k+6+kx-x^2 \\ y' &=& k-2x \quad | \quad \text{The slope of the line is } -2 \\ -2 &=& k-2x \\ \mathbf{k} & \mathbf{=} & \mathbf{ 2x-2} \quad & \quad \mathbf{x=\dfrac{2+k\pm \sqrt{k^2+8k-20 } } {2}} \\\\ k & = & 2+k\pm \sqrt{k^2+8k-20 } -2 \\ \sqrt{k^2+8k-20 } &=& 0 \\ k^2+8k-20 &=& 0 \\ k &=& \dfrac{-8\pm \sqrt{64-4\cdot (-20) } } {2} \\ k &=& \dfrac{-8\pm \sqrt{144} } {2} \\ k &=& \dfrac{-8\pm 12} {2} \\\\ k_1 &=& \dfrac{-8+ 12} {2} \\ \mathbf{k_1} & \mathbf{=} & \mathbf{2} \\\\ k_2 &=& \dfrac{-8- 12} {2} \\ \mathbf{k_2} & \mathbf{=} & \mathbf{-10} \\ \hline \end{array} \)

The two values of k are 2 and -10