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3 years ago

Which inequality is true if p=3.4

Mathematics

1 answer:

Serga [27]3 years ago

7 0

Answer:

None of the inequalities are correct if p = 3.4

Step-by-step explanation:

Mathematical inequality is an order relation proposition existing between two algebraic expressions connected through the signs: unequal than ≠, greater than>, less than <, less than or equal to ≤, as well as greater than or equal to ≥, resulting in both expressions of different values. That is, an inequality is a relationship that exists between two quantities or expressions and, which indicates that they have different values.

To check an inequality, you must replace the inequality variable with the value of the solution. In this case, being p=3.4, you must do it with each of the inequalities, as shown below:

A. 3p<10.2 ⇒ 3*3.4<10.2 ⇒ 10.2 <10.2

This inequality is not true, because 10.2 is not less than 10.2, but they are equal.

B. 13.6<3.9p ⇒ 13.6<3.9*3.4 ⇒ 13.6<13.26

This inequality is not true, because 13.6 is greater than 13.26 and not less as the inequality shows.

C. 5p>17.1 ⇒ 5*3.4 >17.1 ⇒ 17 > 17.1

This inequality is not true, because 17 is less than 17.1 and not greater as the inequality shows.

D. 8.5> 2.5p ⇒ 8.5 >2.5*3.4 ⇒ 8.5 > 8.5

This inequality is not true, because 8.5 is not less than 8.5. These values are the same.

None of the inequalities are correct if p = 3.4

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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

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$