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natulia [17]
3 years ago
12

What is the equation of the midline for the function f (x)? f (t) = 3 cos (x) – 2.5

Mathematics
1 answer:
zaharov [31]3 years ago
6 0

Answer:

  y = -2.5

Step-by-step explanation:

For such a problem as this, you can replace all sine or cosine functions with their midline value of 0. Then you have ...

  f(x) = 0 -2.5

which simplifies to ...

  f(x) = -2.5

You can leave the equation like this, or write it as ...

  y = -2.5

_____

Perhaps you can see that the midline is the value of any constant added to a sine or cosine function.

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The answer to your question is 6.67.
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3 years ago
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A motorist is driving across the country at a steady speed at 54.5 miles per hour. In the 17th hour, the odometer on her car rea
timama [110]

54.5*17=926.5

62416-926.5=61489.5

8 0
3 years ago
Bryce randomly chooses a card from a standard deck of 52 containing 13 of each suit. He records the suit of each card in the tal
Nikolay [14]

Answer:

52 cards  / 4 suits  = 13 cards of each suit.

Theoretically picking a heart would be 13/52 = 1/4 probability.

Experimentally she picked 15 hearts out of 80 total tries. for a 15/80 = 3/16 probability, which is less than the theoretical probability.

1/4 - 3/16 = 1/16

The answer is A.

Step-by-step explanation:

The right answer is A - The theoretical probability of choosing a heart is StartFraction 1 over 16 EndFraction greater than the experimental probability of choosing a heart

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
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No because there has to be a remaining opposite persent
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