The axis of symmetry for the function f(x)=-2x^2+4x+1 is the line x=1 where it's the vertex of the function located

Mathematics

2 answers:

vodomira [7]3 years ago

8 0

Vertex has the coordinate (1;3)

arlik [135]3 years ago

3 0

Answer:

1,3

Step-by-step explanation:

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lets say you start with the equation 12b=24. what step should you take to find the quantity that equals 4b

allsm [11]

Answer:

4b = 8

Step-by-step explanation:

Given 12b = 24

This is written as (4 X 3)b = 24

⇒ 4b X 3 = 24

Dividing by 3 throughout, we have

4b = $\frac{24}{3}$ = 8.

Therefore, the value of 4b is 8.

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3 years ago

Can this visual be described with a linear, exponential, or neither

Pachacha [2.7K]

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Step-by-step explanation:

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3 years ago

Calculate the following limit:

aleksklad [387]

$\lim_{x\to\infty}\dfrac{\sqrt x}{\sqrt{x+\sqrt{x+\sqrt x}}}=\\
\lim_{x\to\infty}\dfrac{\dfrac{\sqrt x}{\sqrt x}}{\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt x}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{\sqrt{x^2}}}}=\\$
$\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{x+\sqrt x}{x^2}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{\sqrt x}{\sqrt{x^4}}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{x}{x^4}}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{1}{x^3}}}}}=\\
=\dfrac{1}{\sqrt{1+\sqrt{0+\sqrt{0}}}}=\\$
$=\dfrac{1}{\sqrt{1+0}}=\\
=\dfrac{1}{\sqrt{1}}=\\
=\dfrac{1}{1}=\\
1
$

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3 years ago

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vichka [17]

Answer:

the value of sin(F) is 3/5

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3 years ago

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kykrilka [37]

The answer is all real numbers for x

because simplifying, you get both sides are equal