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4 years ago

The difference between 2 and twice a number is equal to the sum of a number and 5

Mathematics

1 answer:

kobusy [5.1K]4 years ago

6 0

Answer: 7 or -1

Step-by-step explanation: Let the unknown number be x

Since the question did not say " positive " difference , we could interpret the statement in two ways:

1. 2x - 2 = x + 5

collecting the like terms

2x -x = 5 + 2

x = 7

2. 2 - 2x = x + 5

2 - 5 = x + 2x

-3 = 3x

x = -1

check: for the first one let us confirm if substituting 7 into the equation will give an equal result

2x - 2 = x + 5

2(7) - 2 = 7 + 5

14 - 2 = 12

12 = 12

you can see that they are equal

Let us confirm the other value too, by substituting -1 into the equation

2 - 2x = x + 5

2 - 2 ( -1 ) = -1 + 5

2 + 2 = 4

4 = 4

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3 years ago

Apply Gaussian quadrature with n = 4 to approximate integrate sin x^2dx from 1 to 5

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First, recall that Gaussian quadrature is based around integrating a function over the interval [-1,1], so transform the function argument accordingly to change the integral over [1,5] to an equivalent one over [-1,1].

$x=2t+3\iff t=\dfrac x2-\dfrac32\implies2\mathrm dt=\mathrm dx$
$x=1\implies t=\dfrac{2-6}4=-1$
$x=5\implies t=\dfrac{10-6}4=1$

So,

$\displaystyle\int_{x=1}^{x=5}\sin x^2\,\mathrm dx=\displaystyle2\int_{t=-1}^{t=1}\sin(2t+3)^2\,\mathrm dt$

Let $f(t)=2\sin(2t+3)^2$ . With $n=4$ , we're looking for coefficients $c_i$ and nodes $x_i$ , with $1\le i\le4$ , such that

$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx c_1f(x_1)+\cdots+c_4f(x_4)$

You can either try solving for each with the help of a calculator, or look up the values of the weights and nodes (they're extensively tabulated, and I'll include a link to one such reference).

Using the quadrature, we then have

$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.3749f(-0.8611)+0.6521f(-0.3400)+0.6521f(0.3400)+0.3749f-0.8611)$
$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.5790$