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Paladinen [302]

3 years ago

12

There are several ways to prove that a shape is a parallelogram. Describe the various ways and provide a proof using one of the

methods you describe.

1 answer:

Papessa [141]3 years ago

5 0

A parallelogram is a four sided shape, where opposite sides are parallel.

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An elementary school enrolled 57 students in each grade from kindergarten through sixth. How many students are there in the scho

Luba_88 [7]

Step-by-step explanation:

From kindergarten to sixth grade is seven grades and if for each grade 57 new students get enrolled then 57*7=399 there are 399 students

6 0

3 years ago

Read 2 more answers

In the first game of the year, the modified basketball team made 63.5% of their foul shot free throws. Matthew made all 6 of his

SVETLANKA909090 [29]

So 6 shots equal 25% of the teams total shots(x)
So 6= 0.24(x)
X = 6/0.25
X= 24

If the team made 62. 5% of their shots then they missed 100% - 6.25% = 37.5 %
37.5 % of 24 = 24(0.375) = 9

4 0

3 years ago

Someone help me please and thank you!! ASAP :(

pantera1 [17]

<h3>Answer: Choice D) </h3>

Work Shown:

$(f+g)(x) = f(x)+g(x)\\\\(f+g)(x) = \frac{x-23}{x^2+9x-36}+\frac{1}{x+12}\\\\(f+g)(x) = \frac{x-23}{(x+12)(x-3)}+\frac{1(x-3)}{(x+12)(x-3)}\\\\(f+g)(x) = \frac{x-23+1(x-3)}{(x+12)(x-3)}\\\\(f+g)(x) = \frac{x-23+x-3}{x^2+9x-36}\\\\(f+g)(x) = \frac{2x-26}{x^2+9x-36}\\\\$

We must require that $x \ne -12$ and $x \ne 3$ to avoid having 0 in the denominator. This is why choice D is the answer.

3 0

2 years ago

Which linear inequality is represented by the graph?

Virty [35]

The answer is the third one down because y moves up 1/3 and it’s 2 away from each even number

7 0

2 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

2 years ago