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4 years ago

2x+5y=-6 , wht is the x intercept, what is the y intercept

1 answer:

7 0

These are the xx and yy intercepts of the equation 2x−5y=62x-5y=6.x-intercept: (3,0)(3,0)y-intercept: (0,−65)

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Brain was asked to round 4.44 to the nearest tenth and said it was 4.5 Is he correct?Why or Why not?

Annette [7]

Brian (I think you meant Brian haha..) is incorrect because the tenth place is the number right after the decimal which is a 4. If you are rounding to the tenth place that means that you have to look at the number to the right of that, which is 4. 5 and up means you round up and 4 and down means you round down. This is a 4, so you keep the 4 after the decimal and get rid of the last 4. Therefore, when rounded it would be: 4.4.
Hope this helped! Good luck! :)

5 0

4 years ago

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What's the equation for the line that passes through the given point and is parallel to the graph of the given line (3,-1);y=2x-

ad-work [718]

You should plug the x and y values into the original equations to get your b value. You should get a b value of 7. Your new equation should be y=2x+7

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3 years ago

Is this right please help meh

Mama L [17]

Yes you are correct very nice job!

4 0

3 years ago

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Find two vectors in R2 with Euclidian Norm 1 whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .