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3 years ago

Find all zeros for (2x-1)(x-5)

Mathematics

1 answer:

NeTakaya3 years ago

5 0

To do this set each of the factors to 0:

2x - 1 = 0
2x = 1
x = 1/2

x - 5 = 0
x = 5

Hope this helps!

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3 years ago

Can you help me please

MArishka [77]

Answer:

Step-by-step explanation:

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2 years ago

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The graph of an inverse trigonometric function passes through the point (1, pi/2). Which of the following could be the equation

rodikova [14]

Answer: C) $y=sin^-1 x$

Step-by-step explanation:

Since, the graph of an inverse trigonometric function will pass through the point $(1,\frac{\pi}{2})$ ,

If this point satisfies the function,

For the function $y=cos^{-1} x$

If x = 1

$y=cos^{-1}1=0$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cos^{-1} x$ ,

⇒ The graph of $y=cos^{-1} x$ is not passing through the point $(1,\frac{\pi}{2})$

For the function $y=cot^{-1}x$

If x = 1

$y=cot^{-1}1=\frac{\pi}{4}$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cot^{-1}x$ ,

⇒ The graph of $y=cot^{-1}x$ is not passing through the point $(1,\frac{\pi}{2})$

For the function $y=sin^{-1} x$

If x = 1

$y=sin^{-1}1=\frac{\pi}{2}$

Thus, $(1,\frac{\pi}{2})$ is satisfying function $y=sin^{-1} x$ ,

⇒ The graph of $y=sin^{-1} x$ is passing through the point $(1,\frac{\pi}{2})$ .

For the function $y=tan^{-1}x$

If x = 1

$y=tan^{-1}1=\frac{\pi}{4}$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cos^{-1} x$ ,

⇒ The graph of $y=tan^{-1} x$ is not passing through the point $(1,\frac{\pi}{2})$ .

Hence, Option C is correct.

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3 years ago

X+2x^1/2 -63 solve for x

AysviL [449]

You factor this out. So x+2x^(1/2)-63. Factors into (x^(1/2)+9)(x^(1/2)-7). If you distribute it, you can see it gives you your original equation

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3 years ago