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diamong [38]
2 years ago
15

How many solutions exist for the given equation? 3x 13 = 3(x 6) 1 zero one two infinitely many

Mathematics
1 answer:
romanna [79]2 years ago
8 0

An equation is formed of two equal expressions. There is no solution set of the equation 3x+13 = 3(x+6)+1.

<h3>What is an equation?</h3>

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

The equation can be solved in the following manner,

3x+13 = 3(x+6)+1\\3x+13=3x+18+1\\3x+13=3x+19

Since both the sides of the equation are the same in terms of variable and the constant terms are not the same. Therefore, there is no solution set of the equation 3x+13 = 3(x+6)+1.

Learn more about Equation:

brainly.com/question/2263981

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I did the math but I got -4 so maybe -14
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A baby is weighed on a defective scale. He weighed 13 pounds. The baby weighs exactly 10 pounds. What is the percent error?
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Answer:

30%

Step-by-step explanation:

The answer is off by 30%, as the scale is off by 3 pounds. 3/10=30

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3 years ago
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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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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