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3 years ago

Change 6/1/4%to a fraction

Mathematics

1 answer:

Andreas93 [3]3 years ago

4 0

6. 1
_
. 4
I can't write the response understandable on this screen, I hope this is some use to u.

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Which number line represents the solution set for the inequality 3(8-4x)<6(x-5)?

djverab [1.8K]

we have

$3(8-4x) 54\\ x > (54/18)\\x > 3$

the solution is the interval -------> (3,∞)

therefore

the answer in the attached figure

3 0

3 years ago

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Name the triangles that are classified by angles

Bumek [7]

<span>2. scalene, isosceles, equilateral</span>

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3 years ago

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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

Axis of sym: x =

marta [7]

Answer:

<h2>SEE BELOW</h2>

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

quadratic function
PEMDAS

<h3>let's solve:</h3>

vertex:(h,k)

therefore

vertex:(-1,4)

axis of symmetry:x=h

therefore

axis of symmetry:x=-1

to find the quadratic equation we need to figure out the vertex form of quadratic equation and then simply it to standard form i.e ax²+bx+c=0

vertex form of quadratic equation:

y=a(x-h)²+k

therefore

y=a(x-(-1))²+4
y=a(x+1)²+4

it's to notice that we don't know what a is

therefore we have to figure it out

the graph crosses y-asix at (0,3) coordinates

so,

3=a(0+1)²+4

simplify parentheses:

$3 = a(1 {)}^{2} + 4$

simplify exponent:

$3 = a + 4$

therefore

$a = - 1$

our vertex form of quadratic equation is

y=-(x+1)²+4

let's simplify it to standard form

simplify square:

$y = - ( {x}^{2} + 2x + 1) + 4$

simplify parentheses:

$y = - {x}^{2} - 2x - 1 + 4$

simplify addition:

$y = - {x}^{2} - 2x + 3$

therefore our answer is D)y=-x²-2x+3

the domain of the function

$x\in \mathbb{R}$

and the range of the function is

$y\leqslant 4$

zeroes of the function:

$- {x}^{2} - 2x + 3 = 0$

$\sf divide \: both \: sides \: by \: - 1$

${x}^{2} + 2x - 3 = 0$

$\implies \: {x}^{2} + 3x - x + 3 = 0$

factor out x and -1 respectively:

$\sf \implies \: x(x + 3) - 1(x + 3 )= 0$

group:

$\implies \: (x - 1)(x + 3) = 0$

therefore

$\begin{cases} x_{1} = 1 \\ x_{2} = - 3\end{cases}$

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3 years ago

In a game against the Creek Side

frosja888 [35]

Classify the graph as a linear function, nonlinear function, or relation (non- function) 10 O A. Linear function B.

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3 years ago