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2 years ago

13

Find the value of this expression if x = 5 and x=2 xy^2 / -3

1 answer:

Elina [12.6K]2 years ago

8 0

Answer:

-20/3

Step-by-step explanation:

here x=5 and y=2 then put the value and get answer.

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Consider the problem of finding the line of symmetry and vertex of the quadratic equation f(x) =x^2-8x+15 What is the error in t

Jet001 [13]

We are given the quadratic:

$f(x)=x^2-8x+15$ , with a=1, b=-8, c=15.

We know that the x-coordinate of the vertex, which is the point where the line of symmetry passes through is

$\displaystyle{ -\frac{b}{2a}$ .

Thus, the x-coordinate of the vertex is $-\frac{b}{2a} =-\frac{-8}{2\cdot1}= \frac{8}{2}=4$ .

Thus, the line of symmetry is x=4.

Answer: <span>B. The line of symmetry should have been 4 instead of –4. </span>

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3 years ago

Which best describes the triangle or the triangles, if any, that can be formed with two sides that measure 5 inches and an angle

Arte-miy333 [17]

One isosceles triangle

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3 years ago

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Which function has zeros at x=-2 and x=5?

Licemer1 [7]

Answer:

$x^{2} - 3x - 10$

Step-by-step explanation:

To start, lets form our polynomial factors from our zeroes.

x = -2

x + 2 = 0

x = 5

x - 5 = 0

Now lets put these into factored form by moving them into parenthesis and multiplying them with each other!

(x+2)(x-5)

We can start moving them out! I am going to use FOIL, but you could also use distributive or any other method/property.

x · x = x²

x · -5 = -5x

x · 2 = 2x

2 · 5 = 10

Now lets put our terms into our polynomial function, based on the order of the powers.

$x^{2} - 3x - 10$

We are done! Hope this helps!

7 0

2 years ago

How derivative of this absolute value function is like this!

Daniel [21]

Recall the definition of absolute value.

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

When $x$ ,

$|x| = x \implies \dfrac{d|x|}{dx} = 1$

When $x$ ,

$|x| = -x \implies \dfrac{d|x|}{dx} = -1$

The derivative does not exist at $x=0$ , since the one-sided limits

$\displaystyle \lim_{x\to0^-} f'(x) = -1$

and

$\displaystyle \lim_{x\to0^+} f'(x) = +1$

do not match.

So the derivative of $|x|$ is

$\dfrac{d|x|}{dx} = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \\ \text{undefined} & \text{if } x = 0\end{cases}$

Now we can write this as

$\dfrac{d|x|}{dx} = \dfrac x{|x|} = \dfrac{|x|}x$

since

$x > 0 \implies |x| = x \implies \dfrac{|x|}x = \dfrac xx = 1$

and

$x < 0 \implies |x| = -x \implies \dfrac{|x|}x = -\dfrac xx = -1$

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1 year ago

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Arturiano [62]

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2 years ago

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