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Nadya [2.5K]
3 years ago
7

Factoring trinomials

Mathematics
1 answer:
Tcecarenko [31]3 years ago
4 0

c^22c^2 + 13c +21 \\4c^2 + 13(2)c + 42\\(2c + 6)(2c+7)\\(c+3)(2c+7)


This is a way to factoring trinomials (there exist different equivalent methods).

Multiply the trinomial but the term accompanying  c^2. This is the second line. Then, you could take the square of the 4c^2, ant try to create a factor () () that will correspond to the expression in the second line. That is, we want 4c^2 + 13(2) c + 42 = (2c + ?) (2c + ?)

In ? we put the corresponding numbers that, if we multiply them we will obtain 42, and if we add them we will obtain 13. This numbers are 6 and 7. Then, we have (2c + 6) (2c +7)

The last step is divide by the number that we multipy in the first step.

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Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx  = 1

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y)  and replacing a and b with their respective values, we have

\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx  = 1

Since c is a constant, we can bring it out of the integral sign; to give us

c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx  = 1

Open the bracket

c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx  = 1

Integrate with respect to y

c\int\limits^3_0 {\frac{xy^{2}}{2}  +\frac{xy^{3}}{3} } \, dx (0,3}) = 1

Substitute 0 and 3 for y

c\int\limits^3_0 {(\frac{x* 3^{2}}{2}  +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2}  +\frac{x * 0^{3}}{3})} \, dx = 1

c\int\limits^3_0 {(\frac{x* 9}{2}  +\frac{x * 27}{3} ) - (0  +0) \, dx = 1

c\int\limits^3_0 {(\frac{9x}{2}  +\frac{27x}{3} )  \, dx = 1

Add fraction

c\int\limits^3_0 {(\frac{27x + 54x}{6})  \, dx = 1

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Rewrite;

c\int\limits^3_0 (81x * \frac{1}{6})  \, dx = 1

The \frac{1}{6} is a constant, so it can be removed from the integral sign to give

c * \frac{1}{6}\int\limits^3_0 (81x )  \, dx = 1

\frac{c}{6}\int\limits^3_0 (81x )  \, dx = 1

Integrate with respect to x

\frac{c}{6} *  \frac{81x^{2}}{2}   (0,3)  = 1

Substitute 0 and 3 for x

\frac{c}{6} *  \frac{81 * 3^{2} - 81 * 0^{2}}{2}    = 1

\frac{c}{6} *  \frac{81 * 9 - 0}{2}    = 1

\frac{c}{6} *  \frac{729}{2}    = 1

\frac{729c}{12}    = 1

Multiply both sides by \frac{12}{729}

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8 0
3 years ago
How do you do these two questions?
nignag [31]

Step-by-step explanation:

(a) ∫₋ₒₒ°° f(x) dx

We can split this into three integrals:

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Since the function is even (symmetrical about the y-axis), we can further simplify this as:

= ∫₋₁¹ f(x) dx + 2 ∫₁°° f(x) dx

The first integral is finite, so it converges.

For the second integral, we can use comparison test.

g(x) = e^(-½ x) is greater than f(x) = e^(-½ x²) for all x greater than 1.

We can show that g(x) converges:

∫₁°° e^(-½ x) dx = -2 e^(-½ x) |₁°° = -2 e^(-∞) − -2 e^(-½) = 0 + 2e^(-½).

Therefore, the smaller function f(x) also converges.

(b) The width of the intervals is:

Δx = (3 − -3) / 6 = 1

Evaluating the function at the beginning and end of each interval:

f(-3) = e^(-9/2)

f(-2) = e^(-2)

f(-1) = e^(-1/2)

f(0) = 1

f(1) = e^(-1/2)

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f(3) = e^(-9/2)

Apply Simpson's rule:

S = Δx/3 [f(-3) + 4f(-2) + 2f(-1) + 4f(0) + 2f(1) + 4f(2) + f(3)]

S ≈ 2.5103

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denis23 [38]

Answer:

$8.75

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Selling price of the article = Rs 420

As, S.P > C.P

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⠀⠀⠀= Rs 20

So,

Profit percent = (20/400 × 100)%

⠀⠀⠀⠀⠀⠀⠀⠀ = (20/4) %

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