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3 years ago

I’m gonna need a little help on this

Mathematics

1 answer:

Scrat [10]3 years ago

7 0

Answer:

sorry I don't know just ask to any one

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Which statement correctly finds the interquartile range for the set of data represented by the box plot?

gizmo_the_mogwai [7]

Answer:

Step-by-step explanation:

Median: 9.5

Lower Quartile: 6

Upper Quartile: 11

Interquartile Range: upper quartile - lower quartile = answer

6 0

3 years ago

How many ways can the letters in the word CHEWY be arranged to form a five letter code

MakcuM [25]

Answer:

60 ways;

GOOD LUCK!

Step-by-step explanation:

The word CHEWY has 5 letters, Therefore, it can be arranged in 5!/2! = 120/2 = 60 ways.

5 0

1 year ago

Find the value (64)3/2

faltersainse [42]

(64)^3/2 = (sqrt(64))^3 = 8^3 = 512.

3 0

3 years ago

Make a rational equations with the following requirements

Rufina [12.5K]

Answer:

$f(x)=\dfrac{-50(x-1)(x-2)}{(x-5)(x+5)}$

Step-by-step explanation:

Since f(x) has asymptotes at x = 5 and x = -5, then the denominator of the rational function contains the terms (x - 5) and (x + 5):

$f(x)=\dfrac{?}{(x-5)(x+5)}$

Since f(x) has x-intercepts at x = 2 and x = 1, then the numerator of the rational function contains the terms (x - 2) and (x - 1):

$f(x)=\dfrac{A(x-1)(x-2)}{(x-5)(x+5)}$

Now substitute the point (0, 4) and solve for A:

$f(0)=4\\\\\implies \dfrac{A(0-1)(0-2)}{(0-5)(0+5)}=4\\\\\\\implies -\dfrac{2}{25}A=4\\\\\\\implies A=-50$

So final rational function:

$f(x)=\dfrac{-50(x-1)(x-2)}{(x-5)(x+5)}$

4 0

3 years ago

Find the minimum value of the function f(x)=x^2+5x-6

mash [69]

<h3><u>Explanation</u></h3>

Method 1 (Formula)

$\begin{cases}h = - \frac{b}{2a} \\ k = \frac{4ac - {b}^{2} }{2a} \end{cases}$

The vertex of Parabola is the maximum/minimum point depending on the value of a.

Find Vertex

<u>h-value</u>

$h = - \frac{5}{2(1)} \\ h = - \frac{5}{2}$

<u>k-value</u>

$k = \frac{4(1)( - 6) - {(5)}^{2} }{4(1)} \\ k = \frac{ - 24 - 25}{4} \\ k = \frac{ - 49}{4} \\ k = - \frac{49}{4}$

The minimum value is the value of k. Therefore the minimum value is - 49/4 at x = -5/2.

Method 2 (Derivative)

This is Calculus method. We simply differentiate the function then substitute y' = 0.

Differentiate Function

$f'(x) = 2 {x}^{2 - 1} + {5x}^{1 - 1} - 0 \\ f'(x) = 2x + 5$

Substitute f'(x) = 0

$0 = 2x + 5 \\ - 5 = 2x \\ - \frac{5}{2} = x$

Substitute x = -5/2 in the original equation.

$f(x) = {( - \frac{5}{2} )}^{2} + 5( - \frac{5}{2} ) - 6 \\ f(x) = \frac{25}{4} - \frac{25}{2} - 6 \\ f(x) = \frac{25}{4} - \frac{50}{4} - \frac{24}{4} \\ f(x) = \frac{25}{4} - \frac{74}{4} \\ f(x) = - \frac{49}{4}$

<h3><u>Answer</u><u /></h3>

<u> $\sf{the \: \: minimum \: \: value \: \: is \: \: - \frac{49}{4} \: \: at \: \: x = - \frac{5}{2} }$ </u>

4 0

3 years ago