Identify the correct statement for the given figure.

Mathematics

1 answer:

Ket [755]2 years ago

8 0

It’s either A or D

Hope it helps!!! Good luck

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Which quadratic equation is equivalent to (x^2-1)^2-11(x^2-1)+24=0

Y_Kistochka [10]

Answer:

The correct answer is first option

u² - 11u + 24 = 0

When u = (x² - 1)

Step-by-step explanation:

It is given that,

(x² - 1)² - (x² - 1) + 24 = 0

<u>To find the correct answer</u>

Substitute u = x² - 1

The equation becomes,

u² - 11u + 24 = 0 Where u = (x² - 1)

Therefore the correct answer is first option

u² - 11u + 24 = 0

When u = (x² - 1)

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$\frac{y_{2}- y_{1} }{x_{2}- x_{1} }$

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2 years ago

What's 1/3 of $19.95

ki77a [65]

Answer: 6.65

Explanation: 1995/3= 6.65

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This is a question I have been trying to do for quite a while. My additional mathematics exam is next week and I need help. Anyo

Radda [10]

First of all, let's consider the line, since it's simpler to graph: we draw the endpoints and connect them:

$f(x)=2-\dfrac{3x}{4\pi} \implies f(0)=2,\quad f(2\pi)=\dfrac{1}{2}$

So, you just need to draw the points

$(0,2),\quad \left(2\pi, \dfrac{1}{2}\right)$

As for the trigonometric function, we have to start from the parent function $y=\sin(x)$ and derive the graph of its child function via transformations:

When we multiply the whole function by 2, we stretch the graph vertically. So, the function has still period $2\pi$ , but now it ranges from -2 to 2 instead of from -1 to 1 (amplitude 2)
When we multiply the argument by 2, we compress the function horizontally. So, the new period becomes $\pi$ , and the function makes two complete oscillations from 0 to $2\pi$