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2 years ago

Can some help me I will mark u brilliant

Mathematics

2 answers:

Nadya [2.5K]2 years ago

7 0

Answer:

Let's first try to find the radius before finding the diameter because when you find the radius, you double it to get the diameter.

To find the radius, use the area for a circle formula and plug in the values you know so far, so the formula is;

A = π $r^{2}$

What we know:

1,017 is the area.

Plug that area in our equation:-

A = π $r^{2}$

1,017 = 3.14 $r^{2}$

Solve for r;

÷3.14 ÷3.14

324(approximate) = $r^{2}$

√324 = r

18 = r, the radius is around 18 square meters (do not mistake the unit for some other unit).

<h2>Therefore, your answer is C.</h2><h3>Another alternative way;</h3>

Plug in your options' numbers into the equation;

A = π $r^{2}$

And by looking at the options, we can see the only number here being used is 18, so we use 18 in the formula for the area of a circle to check if it'll result in 1,017 as the area.

Also: Notice if it says diameter or radius when using this solution, you are looking for the radius first.

jeka57 [31]2 years ago

3 0

Answer:

The answer is C

Step-by-step explanation:

The formula of a circle is pie radius ^2. Its asking for the surface area. So I got C because you do pie 18^2 and gets you approximately 1017. Although A also says 18 as the radius, it says meters. After you solve a surface area problem you do square meters or square units. For volume you have to use units cubed or units ^3. Make sure to always do this or your teacher might count you as wrong.

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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

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