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3 years ago

What is the value of x in the equation x+7=−213 x+7=−2

Mathematics

1 answer:

MariettaO [177]3 years ago

6 0

Answer:

Step-by-step explanation:

⅓x + 7 = -2

isolate the x term by subtracting 7 from both sides

⅓x = -9

Multiply both sides by 3

x = -27

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2 years ago

Triangle ABC has side lengths: V6, V2, and 2/2 units.

Morgarella [4.7K]

Answer:

90°, 60°, and 30°

60°.

Step-by-step explanation:

The triangle ABC has side lengths √6, √2, and 2√2 units.

It is clear that the triangle ABC is right triangle because (2√2)² = (√6)² + (√2)² that means the side lengths satisfy the Pythagoras Theorem.

Now, if the angle between hypotenuse (2√2 units) and base (√2 units) is $\theta$ , then

$\cos \theta = \frac{\sqrt{2} }{2\sqrt{2} } = \frac{1}{2}$

Hence, $\theta$ = 60°

Therefore, the three angles of the triangle are 90°, 60°, and 30°.

Now, if the base of the triangle is 16 units, then other two side lengths will also change proportionally to remain the triangle a right triangle.

And in that case the base angle will remain 60°. (Answer)

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4 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:

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

8 0

3 years ago