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3 years ago

Need help with number 17

Mathematics

1 answer:

mrs_skeptik [129]3 years ago

8 0

Answer:

the difference between -3^4 and (-3)^4 is that, for -3 to the power 4, the power only affect 3 and not negative which you get the answer has -81 and for (-3)^4 the power affect everything in blacket which result to the answer as positive 81

Read more about combination at:

brainly.com/question/11732255

#SPJ1

4 0

2 years ago

There are 5 cookies. Sue took 4 how much are left

irina [24]

1 is the answer
5-4=1

3 0

3 years ago

Read 2 more answers

Solve for x: 4 over 5 x + 4 over 3 = 2x x = __________________ Write your answer as a fraction in simplest form. Use the "/" sym

Vesna [10]

4/5x + 4/3 = 2x....there are 2 ways to do this...one with fractions, one without.

with :
4/5x + 4/3 = 2x
4/3 = 2x - 4/5x
4/3 = 10/5x - 4/5x
4/3 = 6/5x
4/3 * 5/6 = x
20/18 = x
10/9 = x

without :
4/5x + 4/3 = 2x....multiply by common denominator of 15
12x + 20 = 30x
20 = 30x - 12x
20 = 18x
20/18 = x
10/9 = x

5 0

3 years ago

Read 2 more answers

Heyyy, pls help! due today!

goldenfox [79]

Answer:

Step-by-step explanation: WELL i cant help lm really sorry

6 0

3 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.

3 0

3 years ago