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3 years ago

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This equation has one solution. 3(x – 2) + 4x = 10(x + 1) What is the solution? A.–5 B.–2 C.StartFraction negative 16 Over 3 End

Fraction D.StartFraction negative 16 Over 7 EndFraction

2 answers:

gregori [183]3 years ago

7 0

Answer:

x = -16/3

Step-by-step explanation:

All work is in the image below!

Alecsey [184]3 years ago

5 0

Answer:

-16/3 = x

Step-by-step explanation:

3(x – 2) + 4x = 10(x + 1)

Distribute

3x -6 +4x = 10x+10

Combine like terms

7x-6 = 10x+10

Subtract 7x from each side

-6 = 3x+10

Subtract 10 from each side

-16 = 3x

Divide by 3

-16/3 = 3x/3

-16/3 = x

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Step-by-step explanation:

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If the function f(x) = -1/2x - 7 were graphed, which of the following would be true?

Fittoniya [83]

ANSWER

D. As the x-values increase, the y-values would decrease

EXPLANATION

The given function is
$y = - \frac{1}{2} x - 7$

The slope of this function is
$m = - \frac{1}{2}$

Since the slope is negative, there is a negative or an inverse relationship between x and y.

As the values of x increases, the values of y decreases.

Also, as the values of x decreases, the values of y increases.

The first option is not true because as x is becoming bigger and bigger negatively, y will become positive.

For instance, when
$x = - 40$
$y = - \frac{1}{2} ( - 40) - 7$

This implies that,

$y = 20 - 7 = 13$

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Option C is also false because, x can be positive since it is the independent variable.

See graph

Therefore, the correct answer is option D.

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3 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

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