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3 years ago

What is slope? Vbxnkdb

Mathematics

2 answers:

Elis [28]3 years ago

7 0

Answer:

slope is m= rise/run

Step-by-step explanation:

slope is how steep the points go on the graph

Artist 52 [7]3 years ago

3 0

Answer:

The slope of a line characterizes the direction of a line. To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points .

Step-by-step explanation:

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7-70. Use your preferred method to solve

kirill [66]

Answer:

You would want to divide 6 by 2, getting x=3.

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2 years ago

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Find value of p in..6p-1/3(p+1)=8(p+1/32

puteri [66]

The answer is,

$p = - \frac{1}{4}$

6 0

2 years ago

Whick error did she do and what the correct answer

marusya05 [52]

Answer:

The second one. The -2y should have been +2y

Step-by-step explanation:

a negative multiplied by a negative gives you a positive

8 0

3 years ago

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What is the area of this figure?

VashaNatasha [74]

Answer:

104cm^2

Step-by-step explanation:

Area of 1st rectangle

given

l=12cm

w=6cm

A =l×w

=12cm×6cm

=72cm^2

Area of 2nd rectangle

given

l=14-6=8cm

w=4cm

A=l×w

=8cm×4cm

=32cm^2

Total area =72cm^2+32cm^2=104cm^2

8 0

2 years ago

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Find two vectors in R2 with Euclidian Norm 1<br> 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 $x_2$ .

Remembering,

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

The two vectors we are looking for are:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

5 0

2 years ago