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devlian [24]
3 years ago
6

If s(x) = x – 7 and t(x) = 4x2 – x + 3, which expression is equivalent to (t*s)(x) ?

Mathematics
2 answers:
Inessa05 [86]3 years ago
8 0
Hello,

(t*s)(x)=t(x)*s(x)=(4x²-x+3)*(x-7)

Answer D
Gemiola [76]3 years ago
4 0
The expression (t*s)(x) may also be written as t(x)*s(x) which means that the functions are just multiply to each other. From the given,
                           t(x)*s(x) = (4x² - x + 3)(x - 7)
Thus, the answer for this item is letter D. 
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1. Make y the subject of the equation
y = -4 - 6x


2. Substitute values for guess and check
8 = -4 - (6 x 2)
8 = -4 - 12
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2 = -4 - (6 x 1)
2 = -4 -6
This is incorrect.

-10 = -4 - (6 x 1)
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The correct answer is C.
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5 0
3 years ago
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PLZ HELP!!! Use limits to evaluate the integral.
Marrrta [24]

Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:

\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

r_i=\dfrac{2i}n

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At these sampling points, the function takes on values of

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We approximate the integral with the Riemann sum:

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3

Recall that

\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4

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\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}

Just to check:

\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28

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