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Amiraneli [1.4K]
3 years ago
3

1. How many centavos are there in P 125.75?

Mathematics
2 answers:
stiks02 [169]3 years ago
8 0

Answer:

<u><em>please mark brainliest!</em></u>

Step-by-step explanation:

15 centavos

valkas [14]3 years ago
3 0

Answer:

3

Step-by-step explanation:

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x=\frac{12}{13}

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natita [175]

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Use a system of equations to find the parabola of the form yequals=axsquared2plus+bxplus+c that goes through the three given poi
Andrei [34K]
Given that a parabola of the form y=ax^2+bx+c passing through the points (2, -11), (-2, -23) and (4, -53)

Thus, substituting the points we have:

-11=(2)^2a+2b+c \\  \\ \Rightarrow-11=4a+2b+c\ .\ .\ .\ (1) \\  \\ &#10;-23=(-2)^2a+(-2)b+c \\  \\ \Rightarrow -23=4a-2b+c\ .\ .\ .\ (2) \\  \\ &#10;-53=(4)^2a+4b+c \\  \\ \Rightarrow-53=16a+4b+c\ .\ .\ .\ (3)

We solve equations (1), (2) and (3) simulataneously. (There are many mays it can be solved but I will use row reduction method here).

We form the augumented matrix for equations (1), (2) and (3) and perform elementary row operations as follows:

\left[\begin{array}{ccc}4&2&1\\4&-2&1\\16&4&1\end{array}\right|  \left.\begin{array}{c}-11\\-23\\-53\end{array}\right]  \ \ \ \ \  \frac{1}{4} R_1\rightarrow R_1 \\  \\   \left[\begin{array}{ccc}1& \frac{1}{2} & \frac{1}{4} \\4&-2&1\\16&4&1\end{array}\right|  \left.\begin{array}{c}- \frac{11}{4} \\-23\\-53\end{array}\right]  \ \ \ \ \  {{-4R_1+R_2\rightarrow R_2} \atop {-16R_1+R_3\rightarrow R_3}}

\left[\begin{array}{ccc}1& \frac{1}{2} & \frac{1}{4} \\0&-4&0\\0&-4&-3\end{array}\right|  \left.\begin{array}{c}- \frac{11}{4} \\-12\\-9\end{array}\right]  \ \ \ \ \  - \frac{1}{4} R_2  \\  \\  \left[\begin{array}{ccc}1& \frac{1}{2} & \frac{1}{4} \\0&1&0\\0&-4&-3\end{array}\right|  \left.\begin{array}{c}- \frac{11}{4} \\3\\-9\end{array}\right]  \ \ \ \ \   {{ -\frac{1}{2} R_2+R_1\rightarrow R_1} \atop {4R_2+R_3\rightarrow R_3}}

\left[\begin{array}{ccc}1&0& \frac{1}{4} \\0&1&0\\0&0&-3\end{array}\right|  \left.\begin{array}{c}- \frac{17}{4} \\3\\3\end{array}\right]  \ \ \ \ \  - \frac{1}{3} R_3 \\  \\  \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right|  \left.\begin{array}{c}-4\\3\\-1\end{array}\right]  \ \ \ \ \  - \frac{1}{4} R_3+R_1\rightarrow R_1

Thus, a = -4, b = 3, c = -1

Therefore, the required polynomial is y=-4x^2+3x-1
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