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katrin2010 [14]
1 year ago
9

The initial and terminal points of a vector are given. Write the vector as a linear combination of the standard unit vectors i a

nd j.Initial Point Terminal Point(-7, 1) (9, -3)
Mathematics
1 answer:
gtnhenbr [62]1 year ago
8 0
U=16i-4j

Explanation

any vector in the plane can be writen as a linear combination of the 2 standar unit vectors

\begin{gathered} i=\langle1,0\rangle\text{   j=}\langle0,1\rangle \\ as\text{ follows} \\ V=\langle a,b\rangle=a\langle1,0\rangle+b\langle0,1\rangle=ai+bj \end{gathered}

hence

Step 1

component form

\begin{gathered} U=\text{ terminal point -initial point} \\ U=(9-(-7),(-3-1)=\langle16,-4\rangle \end{gathered}

now, as a linear combination

U=16i-4j

I hope this helps you

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What is the solution to the linear equation? (2/3)x-1/2=1/3+(5/6)x
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Answer:

x=-5

Step-by-step explanation:

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At a wedding, each person is estimated to eat 2/3 of a cup of soup. How many people will be fed by 12 cups of soup?
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You’re asking how many times does 2/3 fit into 12. This needs to be set up as 12 divided by 2/3.
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3 years ago
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I need help with 12 13 and 14​
nikdorinn [45]

Answer: Lines \frac{}{BC} and \frac{}{EF} are different lengths.

Step-by-step explanation:

The distance formula is \sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }, and you can use this formula to solve for the lengths of both lines \frac{}{BC} and \frac{}{EF}.

For line \frac{}{BC}, let x_{1} = the x at point B, or 1, and let x_{2} = the x at point C, or 2.

Now, let y_{1} = the y at point B, or 4, and let y_{2} = the y at point C, or -1.

Now, solve the formula to find the length \frac{}{BC} = \sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\.

\frac{}{BC} = \sqrt{(1-2)^{2} +(4-(-1))^2

\frac{}{BC} = \sqrt{(-1)^{2} +(4+1)^2

\frac{}{BC} = \sqrt{1 +5^2

\frac{}{BC} = \sqrt{(1+25)

\frac{}{BC} = \sqrt{26} \\

Now, for line \frac{}{EF}, let x_{1} = the x at point E, or -4, and let x_{2} = the x at point F, or -1.

Let y_{1} = the y at point E, or -3, and let y_{2} = the y at point F, or 1.

Now, solve the formula to find the length \frac{}{EF} = \sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\.

\frac{}{EF} = \sqrt{(-4-(-1))^{2} +(-3-1)^2

\frac{}{EF} = \sqrt{(-4+1)^{2} +(-4)^2

\frac{}{EF} = \sqrt{(-3)^2+16

\frac{}{EF} = \sqrt{(9+16)

\frac{}{EF} = \sqrt{25}

\frac{}{EF} = 5

Now, look back at \frac{}{BC}. The two lines have different lengths, so you have now justified the fact that they are not the same.

Questions 13 and 14 would be solved in much the same way- but please let me know if you want me to show the work for those as well!

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