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alex41 [277]
3 years ago
8

Write an equation that shows the relationship in this table 1,50 2,100 3,150 4,200

Mathematics
1 answer:
konstantin123 [22]3 years ago
5 0

Answer:

x₁ = 15    xₙ = x₁ * 7 * n     n>1

Step-by-step explanation:

150

2100 = 150 x 7 x 2

3150 = 150 x 7 x 3

4200 = 150 x 7 x 4

x₁ = 150

xₙ = x₁ * 7 * n     n>1

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Answer:

it would be 27 units between line A and B

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2 years ago
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baherus [9]

Answer:

divide 500 by 125.

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Answer is 142

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3 0
2 years ago
If f(x)=sinx−1/cos2x, then limx→π2f(x) is equivalent to which of the following?
Harrizon [31]

Answer:

2

Step-by-step explanation:

Given g(x) = sin(x)-1/cos2(x), we are to find the limit if the function g(x) as g(x) tends to π/2

Substituting π/2 into the function

lim x-->π/2 sin(x)-1/cos 2(x)

= sin(π/2) - 1/cos(2)(π/2)

= 1 - 1/cosπ

= 1- 1/-1

= 1 -(-1)

= 1+1

= 2

Hence the limit of the function h(x) = sin(x)-1/cos2(x) as x--> π/2 is 2

8 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
Answer asap plz and thank you a head of time question in pic
ahrayia [7]
I think the answer is B.
3 0
2 years ago
Read 2 more answers
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