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MariettaO [177]

3 years ago

10

What is the slope of this graph? A. −14 B. 4 C.−4 D. 14

2 answers:

DiKsa [7]3 years ago

6 0

What graph are you talking about:? can you give a pivture

Anna007 [38]3 years ago

3 0

I'm sorry for the inconvenience but can you put a picture a visual image for me to look at so I can go into further analysis and answer the question that you have provided. If you are having trouble with doing that, reach out for help and i will be glad to assist you. :)

That would be greatly appreciated for your kindness :)

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Can anyone help me with my homework?

Evgesh-ka [11]

I want more points next time....Hehehee!

<h2><u>______________________</u></h2>

<em>Solution in attachments!</em>

6 0

2 years ago

Convert 550cm to meters

Viefleur [7K]

1 meter = 100 cm

550/100 = 5.5

5.5 meters

4 0

3 years ago

Help me on number 44 plz

Delvig [45]

All you have to do is write a positive number. let's say 4, then write a positive exponent (that's the tiny number in the right corner of the regular number) and then just write a positive exponent less than 4.
So 4 squared

4 0

3 years ago

let x1,x2, and x3 be linearly independent vectors in R^(n) and let y1=x2+x1; y2=x3+x2; y3=x3+x1. are y1,y2,and y3 linearly indep

Nutka1998 [239]

Answer with Step-by-step explanation:

We are given that

$x_1,x_2$ and $x_3$ are linearly independent.

By definition of linear independent there exits three scalar $a_1,a_2$ and $a_3$ such that

$a_1x_1+a_2x_2+a_3x_3=0$

Where $a_1=a_2=a_3=0$

$y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1$

We have to prove that $y_1,y_2$ and $y_3$ are linearly independent.

Let $b_1,b_2$ and $b_3$ such that

$b_1y_1+b_2y_2+b_3y_3=0$

$b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0$

$b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0$

$(b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0$

$b_1+b_3=0$

$b_1=-b_3$ ...(1)

$b_1+b_2=0$

$b_1=-b_2$ ..(2)

$b_2+b_3=0$

$b_2=-b_3$ ..(3)

Because $x_1,x_2$ and $x_3$ are linearly independent.

From equation (1) and (3)

$b_1=b_2$ ...(4)

Adding equation (2) and (4)

$2b_1==0$

$b_1=0$

From equation (1) and (2)

$b_3=0,b_2=0,b_3=0$

Hence, $y_1,y_2$ and $y_3$ area linearly independent.

5 0

3 years ago

Simultaneously x-2y=0 40-y=14

harkovskaia [24]

Answer:

x = 52, y = 26

Step-by-step explanation:

x - 2y = 0

40 - y = 14

(40 - 40) - y = 14 - 40

- y = -26

y = 26

x - 2(26) = 0

x - 52 = 0

x (- 52 + 52) = 0 + 52

x = 52

7 0

3 years ago