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3 years ago

Answer both 5 and 6 and i will give brainlyest

Mathematics

2 answers:

Karo-lina-s [1.5K]3 years ago

6 0

Answer:

Step-by-step explanation:

lawyer are wrong trust me

Zolol [24]3 years ago

5 0

Answer:

5.I believe it is A 6.c

Step-by-step explanation:

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If the divisor is 4 and the quotient is 8, what is the dividend ?

kolbaska11 [484]

<h3>If the divisor is 4 and the quotient is 8, Then, dividend is 32</h3>

<em><u>Solution:</u></em>

Given that,

Divisor = 4

Quotient = 8

To find: dividend

We know that,

$Dividend \div divisor = quotient$

Therefore,

$Dividend \div 4 = 8\\\\Dividend \times \frac{1}{4} = 8\\\\Dividend = 8 \times 4\\\\Dividend = 32$

Thus, dividend is 32

3 0

3 years ago

Elise walks diagonally from one corner of a square plaza to another. Each side of the plaza is 505050 meters.

Dennis_Churaev [7]

Answer= 70.7 meters.

Step-by-step explanation:

We have been given that Elise walks diagonally from one corner of a square plaza to another. Each side of the plaza is 50 meters.

Since we know that diagonal of a square is product of side length of square and . So we will find diagonal of our given square plaza by multiplying 50 by .

Therefore, diagonal distance across the plaza is 70.7 meters.

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

Question: (Graph attached below)

antoniya [11.8K]

16 I just know that it's 16 GL to ya

6 0

3 years ago

Read 2 more answers

Factor using GCF.<br> Show your work!<br> 1) 36a + 18b - 27c

Mice21 [21]

Match them then add

5 0

2 years ago