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1 year ago

4. (a)(i)Show that log4x=2log16x. (ii)Show that log x=3logb³ x. (iii) Show that log₂x=(1+log₂3)logix.

Mathematics

1 answer:

alexdok [17]1 year ago

3 0

that log4x=2log16x. (ii)Show that log x=3logb³ x. (iii) Show that log₂x=(1+log₂3)logix

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Part 10, What is the area of this figure? NO LINKS!!!

Tamiku [17]

Answer:

shaded area = 601 mm²

Step-by-step explanation:

Area of rectangle = Length * Breadth

Area of first section = 12 * 16 = 192 mm²

Area of 2nd middle section = 8 * ( 16 - 10 ) = 48 mm²

Area of the bottom section = 19 * 19 = 361 mm²

Area of total therefore = 192 mm² + 48 mm² + 361 mm²

= 601 mm²

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2 years ago

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Elena's aunt bought her a $150 savings bond when she was born. When Elena is 20 years old, the bond will have earned 105% in int

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The worth of the bond is $307.50.

<h3>What is the worth of the bond?</h3>

The worth of the bond is the sum of the bond when it was bought and the interest earned on the bond.

Worth of the bond = interest rate + value when the bond was bought

Interest = 105% x $150

1.05 x $150 = $157.70

Worth of the bond = $157.70 + $150 = $307.50

To learn more about interest, please check: brainly.com/question/26164549

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2 years ago

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Find the amount after 2 years for 25000 at 10% pa compounded annually and halfyearly

Molodets [167]

Answer:

8358.7

Step-by-step explanation:

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2 years ago

translate this equation. 126 is the product of goran's score and 7..use the variable g to represent gorans score

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7g = 126 is the translation

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3 years ago

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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

3 years ago

4. (a)(i)Show that log4x=2log16x. (ii)Show that log x=3logb³ x. (iii) Show that log₂x=(1+log₂3)logix.​

4. (a)(i)Show that log4x=2log16x. (ii)Show that log x=3logb³ x. (iii) Show that log₂x=(1+log₂3)logix.