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adell [148]
3 years ago
7

Collection of nickels and quarters is worth $2.85. There are three more nickels and quarters. How many nickels and quarters are

there?
Mathematics
1 answer:
alexdok [17]3 years ago
3 0

The pile contains 12 nickels and 9 quarters.

Let <em>x</em> = the number of nickels and <em>y</em> = the number of quarters.

Then, we have two equations:

(1) $0.05x + $0.25y  = $2.85

(2) x = y +3

Substitute the value of <em>x</em> from Equation (2) into Equation (1).

0.05(<em>y</em>+3) + 0.25<em>y</em> = 2.85

0.05<em>y</em> + 0.15 + 0.25<em>y</em> = 2.85

0.30<em>y</em> = 2.85 – 0.15 = 2.70

y = 2.70/0.30 = 9

Substitute the value of <em>y</em> in Equation (2).

x = 9 + 3 = 12

The pile contains 12 nickels and 9 quarters.

<em>Check</em>: 12×$0.05 + 9×$0.25 = $0.60 + $2.25 = $2.85.

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Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.
Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]

The left and right endpoints of the i-th subinterval, respectively, are

\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8

r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8

for 1\le i\le4, and the respective midpoints are

m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8

  • Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}

  • Midpoint rule

We approximate the area for each subinterval by

M_i=f(m_i)(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}

  • Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial p_i(x), where

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It so happens that the integral of p_i(x) reduces nicely to the form you're probably more familiar with,

S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))

Then the integral is approximately

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0
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HELPPP
Alexxandr [17]
Hope this helps you

3 0
3 years ago
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