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Anna [14]
3 years ago
13

V^2=12^2+2(-6)(9) answer please i am waitind

Mathematics
2 answers:
il63 [147K]3 years ago
3 0

Answer:

V = 6 or V = -6

Step-by-step explanation:

Solve for V over the real numbers:

V^2 = 12^2 - 6 2 9

12^2 + 2 (-6) 9 = 36:

V^2 = 36

Take the square root of both sides:

Answer: V = 6 or V = -6

ExtremeBDS [4]3 years ago
3 0

Answer:

v=6 or v=−6

Step-by-step explanation:

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Let ​f(x)=x^2+6x−16​.<br> Enter the x-intercepts of the quadratic function in the boxes.
Sauron [17]

\underline{\underline{\large\bf{Given:-}}}

\red{\leadsto}\:\textsf{}\sf f(x)= x^2+6x−16

\underline{\underline{\large\bf{To Find:-}}}

\orange{\leadsto}\:\textsf{ x-intercepts of the quadratic function }\sf

\\

\underline{\underline{\large\bf{Solution:-}}}\\

\longrightarrowx- intercept is the point where graph of given function touches the x-axis,f(x) becomes 0 at the point where graph of given function touches the x-axis. Therefore, we would to solve x^2+6x−16=0 and find its root.

\begin{gathered}\\\implies\quad \sf x^2+6x−16=0 \\\end{gathered}

\begin{gathered}\\\implies\quad \sf x^2+8x-2x −16=0 \\\end{gathered}

\begin{gathered}\\\implies\quad \sf x(x+8)-2(x+8)=0 \\\end{gathered}

\begin{gathered}\\\implies\quad \sf (x-2)(x+8)=0\\\end{gathered}

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Need help pleaseI was bad at math in school so lwant to learn
aleksklad [387]

The probability of an event is expressed as

Pr(\text{event) =}\frac{Total\text{ number of favourable/desired outcome}}{Tota\text{l number of possible outcome}}

Given:

\begin{gathered} \text{Red}\Rightarrow2 \\ \text{Green}\Rightarrow3 \\ \text{Blue}\Rightarrow2 \\ \Rightarrow Total\text{ number of balls = 2+3+2=7 balls} \end{gathered}

The probability of drwing two blue balls one after the other is expressed as

Pr(\text{blue)}\times Pr(blue)

For the first draw:

\begin{gathered} Pr(\text{blue) = }\frac{number\text{ of blue balls}}{total\text{ number of balls}} \\ =\frac{2}{7} \end{gathered}

For the second draw, we have only 1 blue ball left out of a total of 6 balls (since a blue ball with drawn earlier).

Thus,

\begin{gathered} Pr(\text{blue)}=\frac{number\text{ of blue balls left}}{total\text{ number of balls left}} \\ =\frac{1}{6} \end{gathered}

The probability of drawing two blue balls one after the other is evaluted as

\begin{gathered} \frac{1}{6}\times\frac{2}{7} \\ =\frac{1}{21} \end{gathered}

The probablity that none of the balls drawn is blue is evaluted as

\begin{gathered} 1-\frac{1}{21} \\ =\frac{20}{21} \end{gathered}

Hence, the probablity that none of the balls drawn is blue is evaluted as

\frac{20}{21}

8 0
1 year ago
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