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3 years ago

What is the monomial if a square of a monomial is: c 0.0001a^8x^4

Mathematics

1 answer:

konstantin123 [22]3 years ago

3 0

Answer:

The answer to your question is 0.1a⁴x²

Step-by-step explanation:

0.0001a⁸x⁴

Process

1.- Get the square root of 0.0001

$\sqrt{0.0001} = 0.1$

2.- Get the square root of a⁸

a⁸ = (a⁴)² $\sqrt{a^{8}} = a^{4}$

3.- Get the square root of x⁴

x⁴ = (x²)² $\sqrt{x^{4}} = x^{2}$

4.- Write the solution

0.1a⁴x²

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Spoilers for answer:
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3 years ago

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3 years ago

The total cost for tuition plus room and board at State University is $2,584.00. Tuition costs $100 more than twice room and boa

liq [111]

Answer:

$1756

Step-by-step explanation:

1. More = Addition

2. Twice = Multiplication by 2

3. Tuition costs $100 more than twice room and board

Tuition = 2x + 100

$2584 = (2x + 100) + x

$2584 = 3x + 100

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7 0

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.

3 0

2 years ago

shusha [124]

Step-by-step explanation:

-4x + 3 < 23

-4x < 20

x > -5

hmm,I don't think we need to put equal sign instead of (<) , for when we started calculating.

we can just keep the question as it is and calculate.

3 0

2 years ago

Read 2 more answers