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What is the value of x? *
4x + 32
172°
Your answer

Answers

Answer 1

Answer:

Answer:

x=35

Step-by-step explanation:

Because the 2 lines are parallel u kno 172=4x+32

from there: 172-32=4x or 140=4x, and then 140/4=x

Answer 2

Answer:

Final answer:

To solve for 'x' in the equation 4x + 32 = 172, we isolate 'x' by first subtracting 32 from both sides of the equation. This gives us 4x = 140. Then, solving for 'x', we divide both sides by 4, resulting in 'x' equal to 35.

Explanation:

To find the value of x, we need to use the process of algebraic simplification. In the equation provided, isolate x by subtracting 32 from both sides of the equation:

4x + 32 = 172

4x = 172 - 32

4x = 140

Next, solve for x by dividing both sides of the equation by 4:

x = 140 / 4

x = 35

Therefore, the value of x in the equation is 35.

Learn more about Algebra here:

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If 40% is eliminated from 500 what is left over?

Answers

Answer:

300

Step-by-step explanation:

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.) π/2 0 3 1 + cos(x) dx, n = 4

Answers

Split up the integration interval into 4 subintervals:

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

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

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

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

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

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

Trapezoidal rule

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

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

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

$p_i(x)=f(\ell_i)((x-m_i)(x-r_i))/((\ell_i-m_i)(\ell_i-r_i))+f(m)((x-\ell_i)(x-r_i))/((m_i-\ell_i)(m_i-r_i))+f(r_i)((x-\ell_i)(x-m_i))/((r_i-\ell_i)(r_i-m_i))$

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.

Final answer:

The question is asking to approximate the definite integral of 1 + cos(x) from 0 to π/2 using the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule for n=4. These are numerical methods used for approximating integrals by estimating the area under the curve as simpler shapes.

Explanation:

This question asks to use several mathematical rules, specifically the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule, to approximate the given integral with a specified value of n which is 4. The integral given is the function 1 + cos(x) dx from 0 to π/2. Each of these rules are techniques for approximating the definite integral of a function. They work by estimating the region under the graph of the function and above the x-axis as a series of simpler shapes, such as trapezoids or parabolas, and then calculating the area of these shapes. The 'dx' component represents a small change in x, the variable of integration. The cosine function in this integral is a trigonometric function that oscillates between -1 and 1, mapping the unit circle to the x-axis. The exact solution would require calculus, but these numerical methods provide a close approximation.