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hjlf
3 years ago
8

Identify the vertex for the graph of y = −3x2 + 6x − 4. (1, 5) (1, −1) (−1, −13) (−1, −9)

Mathematics
2 answers:
mel-nik [20]3 years ago
5 0

\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-3}x^2\stackrel{\stackrel{b}{\downarrow }}{+6}x\stackrel{\stackrel{c}{\downarrow }}{-4} \qquad \qquad  \left(-\cfrac{ b}{2 a}~~~~ ,~~~~  c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{6}{2(-3)}~~,~~-4-\cfrac{6^2}{4(-3)}  \right)\implies \left(1~~,~~-4+\cfrac{36}{12}  \right) \\\\\\ (1~~,~~-4+3)\implies (1~,-1)

Nutka1998 [239]3 years ago
3 0

Answer: (1,-1)

Step-by-step explanation:

Took the test. Got i right

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Put in the values

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3 years ago
Modeling Exponential Growth and Decay In Exercises 119 and 120, Find the exponential function
Lubov Fominskaja [6]

Answer:

There is no exponential function passing through (1,1) and (5,5).

Step-by-step explanation:

We have the following exponential function

y = Ce^{kt}

The function passes through these two points:

(1,1): This means that when t = 1, y = 1

(5,5): This means that when t = 5, y = 5.

So

(1,1)

y = Ce^{kt}

Ce^{k} = 1

e^{k} = \frac{1}{C}

(5,5)

y = Ce^{kt}

5Ce^{k} = 1

Ce^{k} = \frac{1}{5}

From above, we have that:

e^{k} = \frac{1}{C}

C\frac{1}{C} = \frac{1}{5}

1 = \frac{1}{5}

1 cannot be equal to 1/5, so this is wrong.

This means that there is no exponential function passing through (1,1) and (5,5).

5 0
3 years ago
Is X=3 calculate the value of X3+X btw the 3 is a small three
Naddik [55]
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3 years ago
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Given mean of 8.6 words and a standard deviation of 4.3 words. If a random sample of 39 messages, hence, for x > 10:

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