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galben [10]
3 years ago
8

How to determine if set of vectors is linearly independent?

Mathematics
1 answer:
zimovet [89]3 years ago
7 0
The vectors \{v_1,v_2,...,v_p\}\in \mathbb{R}^n are linearly independent if the ONLY solution to the following equation:

c_1v_1+c_2v_2+...c_pv_p=0

is c_1=c_2=...=c_p=0
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5 qt = ? gal????????​
strojnjashka [21]

Answer:

1.25 gal

Step-by-step explanation:

5/4 = 1.25

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3 years ago
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Which number has a value between -8.423 and -8.395?
pav-90 [236]

Answer:

Step-by-step explanation:

Jmeu eimejrimr

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3 years ago
Eighty percent of the tickets for the baseball game on july 4th been sold. there are 20,000 tickets. how many been sold
Viktor [21]
16000 tickets have been sold.

Glad I could help :)
6 0
2 years ago
write the expressions for when you translate the graph of y = |x+2| a) one unit up b) one unit down c) one unit to the left d) o
irina1246 [14]

Hello!

This question is about which values you are changing when you are transforming an equation.

Let's go through the parent function for an absolute value equation and its various transformations.

y=a|b(x-c)|+d

Since we are only looking at horizontal and vertical transformations, we only need to worry about the c and d values.

The c value of a function determines a function's horizontal position, and the d value of a function determines a function's vertical position.

One thing to note here is that the c value is being subtracted from the x value, meaning that if the function is being transformed to the right, you would actually be subtracting that value, while the d value behaves like a normal value, if it is being added, the function is transformed up, and vice versa.

Now that we know this, let's write each expression.

a) y=|x+2|+1

b) y=|x+2|-1

c) y=|x+3|

d) y=|x+1|

Hope this helps!

6 0
2 years ago
Find the Y-coordinate of point P that lies 1/3 along segment RS, where R (-7, -2) and S (2, 4).
QveST [7]

Solution:

Given that the point P lies 1/3 along the segment RS as shown below:

To find the y coordinate of the point P, since the point P lies on 1/3 along the segment RS, we have

\begin{gathered} RP:PS \\ \Rightarrow\frac{1}{3}:\frac{2}{3} \\ thus,\text{ we have} \\ 1:2 \end{gathered}

Using the section formula expressed as

[\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}]

In this case,

\begin{gathered} m=1 \\ n=2 \end{gathered}

where

\begin{gathered} x_1=-7 \\ y_1=-2 \\ x_2=2 \\ y_2=4 \end{gathered}

Thus, by substitution, we have

\begin{gathered} [\frac{1(2)+2(-7)}{1+2},\frac{1(4)+2(-2)}{1+2}] \\ \Rightarrow[\frac{2-14}{3},\frac{4-4}{3}] \\ =[-4,\text{ 0\rbrack} \end{gathered}

Hence, the y-coordinate of the point P is

0

8 0
1 year ago
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