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1 year ago

If the equation below is solved by graphing, which statement is true? log (6 x + 10) = log 1/2 x

Mathematics

1 answer:

kicyunya [14]1 year ago

6 0

The solution to the given expression is x = -20/11

<h3>
What are logarithmic functions?</h3>

Logarithmic function are inverse of exponential functions. Given the equation below;

log (6 x + 10) = log 1/2 x

In order to determine the solution to the given logarithmic equation, we will first have to cancel the logarithm on both sides to have

6x + 10 = 1/2x

Collect the like terms

6x - 1/2x = 0 - 10

Find the LCD

12x-x/2 = -10
11x/2 = -10

Cross multiply

11x = -2 * 10

11x = -20

Divide both sides by 11

11x/11 = -20/11

x = -20/11

Hence the solution to the given expression is x = -20/11

Learn more on log functions and graph here: brainly.com/question/2086094

#SPJ1

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Which is closer to 7?

Anna [14]

Answer:

A. 5.3

Step-by-step explanation:

While both of these numbers are close to 7, 5.3 is the closest and we can find this out with subtraction. So, 9.9 - 7 = 2.9, but 7 - 5.3 = 1.7. Which one can we determine is closer? Well, 2.9 > 1.7, so our answer must be A. 5.3!

Have a nice day and I hope this helped!

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3 years ago

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JulsSmile [24]

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Step-by-step explanation:

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2 years ago

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.

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max2010maxim [7]

You have to use subtraction because it says how many

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Ymorist [56]

The correct answer is B)4

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