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Varvara68 [4.7K]
3 years ago
6

If

Mathematics
2 answers:
algol [13]3 years ago
6 0

Answer:

0

Step-by-step explanation:

We have the two functions:

f(x)=\sqrt{x}+12\text{ and } g(x)=2\sqrt{x}

And we want to find (f-g)(144).

This is the same thing to f(144)-g(144).

So, let’s determine the value of f(144) and g(144) first:

\begin{aligned} f(144)&=\sqrt{144}+12 \\ &=12+12 \\ &=24\end{aligned}

And:

\begin{aligned} g(144)&=2\sqrt{144} \\ &=2(12) \\ &=24 \end{aligned}

Hence:

\begin{aligned} (f-g)(144) &= f(144)-g(144) \\ &=24-24 \\ &=0 \end{aligned}

Our final answer is 0.

Natalija [7]3 years ago
6 0

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

(f - g)(x) =  \sqrt{x}  + 12 - 2 \sqrt{x}  \\

(f - g)(x) =  -  \sqrt{x}  + 12

(f - g)(144) =  -  \sqrt{144}  + 12

(f - g)(144) =  -  \sqrt{ {12}^{2} } + 12

(f - g)(144) =  - (12) + 12

(f - g)(144) =  - 12 + 12

(f - g)(144) = 0

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

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svp [43]

<u>Answer:  </u>

Sum of the roots of the polynomial x^{3}+2 x^{2}-11 x-12 \text { is }-2

<u>Solution:</u>

The general form of cubic polynomial is a x^{3}+b x^{2}+c x+d=0 ---- (1)

If we have any cubic polynomial a x^{3}+b x^{2}+c x+d=0 having roots \alpha , \beta , \theta

Sum of roots \alpha + \beta + \theta = \frac{-b}{a} ---(2)

From question given that,

x^{3}+2 x^{2}-11 x-12 --- (3)

On comparing equation (1) and (3), we get a = 1, b = 2, c = -11 and d = -12

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= \frac{-2}{1}

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Harlamova29_29 [7]

Answer:

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Step-by-step explanation:

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Step-by-step explanation: hope this helps

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