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Taya2010 [7]
3 years ago
12

Plsss help with 4,5,6, or 7

Mathematics
1 answer:
kherson [118]3 years ago
6 0

Answer:

you did not say which question to answer

Step-by-step explanation:

You might be interested in
Ope
ad-work [718]

The slope of the line is 4.83 and the y-intercept is 51

<h3>How to determine slope and the y-intercept?</h3>

From the graph, we can see that the line of the best fit touches the y-axis at y = 51

This means that:

The y-intercept is 51

The slope is then calculated as:

m = (y2 - y1)/(x2 - x1)

Using the points on the line, we have:

m = (80 - 51)/(6 - 0)

Evaluate

m = 4.83

Hence, the slope of the line is 4.83


Read more about line of best fit at:

brainly.com/question/25226042

#SPJ1

6 0
2 years ago
What special characteristics did mahavira share with paul mcCartney?
tia_tia [17]

Answer:

both vegetarians

Step-by-step explanation:

8 0
3 years ago
All rectangles are quadrilaterals.<br><br><br> TrueFalse
mariarad [96]

Answer:

yep

Step-by-step explanation:

yes always

5 0
3 years ago
Read 2 more answers
I would like to copy to the printer is math problem. -4x-7-3x+4=25
Vera_Pavlovna [14]
<span>-4x-7-3x+4=25 then
-7x  -3 = 25 

add 3 to both sides

-7x - 3 + 3 = 25 +3
-7x = 28

Divided both sides by (-7)

-7x/-7 = 28/-7
x = -4

</span>
6 0
3 years ago
Read 2 more answers
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:

  1. v_1 = (x_1,y_1)
  2. 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
3 years ago
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