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3 years ago

Whats the slope of (-3,4) (0, 0)

Mathematics

1 answer:

Maurinko [17]3 years ago

4 0

Answer:

slope = $-\frac{4}{3}$

Step-by-step explanation:

Use the slope formula, $m = \frac{y_2-y_1}{x_2-x_1}$ , to find the slope. $m$ represents the slope. $x_1$ and $y_1$ represent the x and y values of one point, and $x_2$ and $y_2$ represent the x and y values of another point. So, substitute the x and y values of the points (-3, 4) and (0, 0) into the formula appropriately and solve:

$m= \frac{(0)-(4)}{(0)-(-3)} \\m = \frac{0-4}{0+3} \\m = \frac{-4}{3}$

Thus, the slope is $-\frac{4}{3}$ .

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I can't figure out how to do (i + j) x (i x j)for vector calc

Vinil7 [7]

In three dimensions, the cross product of two vectors is defined as shown below

$\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}$

Then, solving the determinant

$\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}$

In our case,

$\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}$

Where we used the formula for AxB to calculate ixj.

Finally,

$\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}$

Thus, (i+j)x(ixj)=i-j

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Answer:

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Step-by-step explanation:

Put x-3 where x is in the original function definition, then "simplify". I think you'll find it convenient to rewrite the original function definition first.

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