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2 years ago

Need help what are the for

Mathematics

2 answers:

sesenic [268]2 years ago

7 0

Answer:

4 and 5

Step-by-step explanation:

stiv31 [10]2 years ago

5 0

Hope you have an amazing day .)

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Let r and s be the roots of x^2 - 6x + 2 = 0. Find (r - s)^2.

Sphinxa [80]

Answer: $(r-s)^{2} = 28$

Step-by-step explanation:

$x^{2} -6x+9=-2+9$
$(x-3)^{2}=7$
$(x-3)= -\sqrt{7}$ or $\sqrt{7}$ <-- there's no plus/minus sign.

$(x-3) = -\sqrt{7}$
$x = 3 -\sqrt{7}$ <-- root.

$(x-3) = \sqrt{7}$
$x = 3+\sqrt{7}$ <-- root.

$[(3-\sqrt{7})-(3+\sqrt{7})]^{2}$
$(3-\sqrt{7} -3 - \sqrt{7})^{2}$ <-- don't forget to distribute the negative.
$(-2\sqrt{7})^{2}$
$(4)(7)$
$28$

4 0

2 years ago

What can you say about the y-help urgent!!!

weeeeeb [17]

The correct answer is ( A )

let me know if you need explanation.

3 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

Please help! What is 1 1/2+1/6?<br><br> -giving out 20 points

Mice21 [21]

$1 \frac{2}{3}$

Hope it helps

7 0

2 years ago

Read 2 more answers

I'm really lost?? help please<br>

BartSMP [9]

If u are lost then go to police and ask for help

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3 years ago