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horsena [70]
3 years ago
6

Help me thanks please please thanks

Mathematics
1 answer:
enot [183]3 years ago
3 0

Answer:

Area of big circle: 2463.01

Area of little circle: 615.75

Area of shaded space: 1,231.51

Step-by-step explanation:

thats what i came up with

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CAN SOMEONE PLEASE HELP ME I NEED THIS ASAP
Fittoniya [83]
The answer is y=-2/2+3
8 0
2 years ago
During practice, the Northwood football team drinks
elena-s [515]

Answer:

The third option (C) Yes, because there is enough water in the cooler for about 81 cups total.Step-by-step explanation:

5 0
3 years ago
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If I buy a stereo for $245 and this includes GST of 10%, what is the pre-GST price?
Ivan

Answer:  $222.73

===========================================

Work Shown:

x = pre-GST price

10% of x = 0.10x = tax amount

x + 0.10x = 1.10x = post-GST price = 245

1.10x = 245

x = 245/1.10

x = 222.7272 approximately

x = 222.73 is the price before tax.

------------

Check:

10% of 222.73 = 0.10*222.73 = 22.273 = 22.27

The tax amount ($22.27) is added to the pre-GST price to get

22.27+222.73 = 245

which matches the post-GST price mentioned.

The answer is confirmed.

Or another way to confirm the answer is to calculate this

1.10*222.73 = 245.003 = 245

6 0
2 years ago
Read 2 more answers
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
Units are squared when dealing with area but not when dealing with perimeter
Mashcka [7]
The answer is true I believe 
7 0
4 years ago
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