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Nutka1998 [239]
2 years ago
5

The parabola with equation $y=ax^2+bx+c$ is graphed below:

Mathematics
2 answers:
sineoko [7]2 years ago
8 0

Answer:

2

Step-by-step explanation:

Source: AOPS

Mashutka [201]2 years ago
4 0

Answer:

m-n=2

Step-by-step explanation:

Instead of using the standard form, we can use the vertex form of a quadratic equation:

f(x)=a(x-h)^2+k

Where a is the leading coefficient, and (h, k) is our vertex.

Our vertex point is at (2, -4). So, let’s substitute 2 for h and -4 for k:

f(x)=a(x-2)^2-4

Now, we need to determine a.

We know that it passes through the point (4, 12). So, when x is 4, y must be 12. In other words:

12=a((4)-2)^2-4

Solve for a. Subtract within the parentheses:

12=a(2)^2-4

Add 4 to both sides:

16=a(2)^2

Square:

16=4a

Solve:

a=4

Thererfore, the value of a is 4.

So, our function is:

f(x)=4(x-2)^2-4

Now, let’s find our roots. Set the equation to 0 and solve for x:

0=4(x-2)^2-4

4=4(x-2)^2\\1=(x-2)^2\\x-2=\pm1 \\ x=2\pm1 \\ x=3\text{ or } 1

So, our roots are 1 and 3.

The greater root is 3 and the lesser root is 1.

Therefore, m-n, where m>n, is 3-1 or 2.

Our final answer is 2.

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Mice21 [21]

Given:

The number is \sqrt{-11}.

To find:

The a+bi form of given number.

Solution:

We have,

\sqrt{-11}

It can be written as

\sqrt{-11}=\sqrt{-1\times 11}

\sqrt{-11}=\sqrt{-1}\times \sqrt{11}      [\because \sqrt{ab}=\sqrt{a}\sqrt{b}]

\sqrt{-11}=i\times \sqrt{11}      [\because \sqrt{-1}=i]

\sqrt{-11}=\sqrt{11}i

Here, real part is missing. So, it can be taken as 0.

\sqrt{-11}=0+\sqrt{11}i

So, a = 0 and b=\sqrt{11}.

Therefore, the a+bi form of given number is 0+\sqrt{11}i.

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2 years ago
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<u>Answer:</u>

<h2>A.</h2>

The graph shows the solution set of the equality √x < -1

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DaniilM [7]

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Step-by-step explanation:

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