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2 years ago

9

F(x)= x^2 +4 +3 /x^2-x-2 not continuous at x = -1.

1 answer:

Ymorist [56]2 years ago

6 0

Answer:

Thats pretty hard man

Step-by-step explanation:

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What is 24.79 x 4,022 pls

lozanna [386]

Answer:

$24.79 \times 4022 \\ = 99705.38$

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2 years ago

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PLS ANSWER PLS PLS PLS

s2008m [1.1K]

The correct answer is 120 degrees.

With the bigger triangle you can add all of the angles to get 60 degrees. (35+25)

This leaves you with a 120 degrees angle in the unmarked corner of the triangle. Any horizontal line and any triangle is 180 in total.

So with a 120 degree angle in the corner this leaves us with a 60 angle in the smaller triangle. ( The smaller triangle equals 60 degrees in all corners, it is equalateral ).

X is on a horizontal line and we now know that the one side of the line equals 60 degrees and any horizontal line equals 180 in total, this means that the measure of X is 120, (60+120)

X is 120 :)

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2 years ago

2(v + 1). Simplify the expression

serg [7]

Answer:

2v+2

Step-by-step explanation:

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2 years ago

How do we calculate 25% of 300

Natali5045456 [20]

25% is equal to $\frac{1}{4}$ . To get 25% of 300, we multiply 300 by either 0.25 or $\frac{1}{4}$ . This gives us $(300)( \frac{1}{4}) = \frac{300}{4} = \frac{150}{2} = 75$

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3 years ago

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Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

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

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

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

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 $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

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3 years ago