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Karolina [17]
2 years ago
8

I’m failing pls help I’ll brainlest ASAP

Mathematics
1 answer:
GuDViN [60]2 years ago
7 0
-2,-1
step by step explanation: i counted the numbers and got the answer
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Which of the following would best represent a cosine function with an amplitude of 2, a period of pi over 3, and a midline at y
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C. f(x) = – 2 cos 6x + 1

Start by determining the amplitude. Since we've deduced the amplitude is 2, the equation can include either a positive or negative 2 (since amplitude measures absolute value).

Next is the period. The equation for period P is P = (2pi)/b. If P is pi/3, then
pi/3 = (2pi)/b. Thus your b value should be 6.

Finally, the midline would be given by + 1 since adding a unit shifts the function upwards. This means that instead of the highest y value being 2 and the lowest -2, instead you'd have values of 3 and -1.
(3 – 1)/2 = 1 (midpoint theory).
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Please help me with this
Ne4ueva [31]

Answer:

Option c

Step-by-step explanation:

A system of equations are given to us. And we need to solve them . The given system is

\begin{cases} y = 2x - 3\dots 1 \\ y = x^2 - 3\dots 2 \end{cases}

We numbered the equations here . Now put the value of equation 1 in equation 2 that is substituting y = 2x - 3 in eq. 2 .

\implies y = x^2 - 3 \\\\\implies 2x - 3 = x^2 - 3 \\\\\implies x^2 - 2x = 0 \\\\\implies x(x-2) = 0 \\\\\implies\red{ x = 0 , 2 }

We got two values of x as 0 & 2 . Alternatively substituting these values we have ,

\implies y = 2 x - 3 \\\\\implies y = 2(0)-3 \qquad or \qquad y = 2(2)-3 \\\\\implies y = 0-3 \qquad or \qquad 4 - 3 \\\\\implies \red{ y = -3 , 1 }

Thefore the required answer is ,

\red{Option\:c} \begin{cases} (0,-3) \\ (2,1) \end{cases}

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