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Triss [41]
3 years ago
5

Write the ordered pair that represents YZ. Then find the magnitude of YZ. y(-4,12), Z(1,19).

Mathematics
2 answers:
ki77a [65]3 years ago
6 0

The magnitude of YZ is 8.6

<u>Explanation:</u>

<u />

Y( -4, 12)

Z ( 1, 19)

Magnitude of YZ = ?

We know:

YZ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

On substituting the value we get:

YZ = \sqrt{(1+4)^2 + (19 - 12)^2} \\\\YZ = \sqrt{(5)^2 + (7)^2} \\\\YZ = \sqrt{25+49} \\\\YZ = \sqrt{74} \\\\YZ = 8.6

Thus, the magnitude of YZ is 8.6

Maurinko [17]3 years ago
3 0

Answer:

its <5,7> \sqrt{74}

Step-by-step explanation:

when you put it into edge, you can write it as <5,7> sqrt 74

and it will mark it correct.

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0.2, 0.6,0.25,0.75,0.82

Step-by-step explanation:

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3 0
4 years ago
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hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and
Mariulka [41]

Given:

The equation is,

2\log _3x-\log _3(x-2)=2

Explanation:

Simplify the equation by using logarthimic property.

\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{      \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}

Simplify further.

\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}

Solve the quadratic equation for x.

\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

\begin{gathered} x-6=0 \\ x=6 \end{gathered}

For (x - 3) = 0,

\begin{gathered} x-3=0 \\ x=3 \end{gathered}

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0
1 year ago
11 –2<br><br> · <br> 3 –4<br><br> · <br> 6 2<br><br> =
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Answer: negative 243

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