Let's consider an arbitrary 2x2 matrix as an example,

The columns of

are linearly independent if and only if the column vectors

are linearly independent.
This is the case if the only way we can make a linear combination of

reduce to the zero vector is to multiply the vectors by 0; that is,

only by letting

.
A more concrete example: suppose

Here,

and

. Notice that we can get the zero vector by taking

and

:

so the columns of

are not linearly independent, or linearly dependent.
Answer: who knows what the answers is bruh
Step-by-step explanation:
I'm taking the liberty of editing your function <span>v = e5xey: It should be
</span>
<span>v = e^5x^ey, with " ^ " indicating exponentiation.
</span>
Did you mean e^(5x) or (e^5)x? I'll assume it's e^(5x).
The partial of v = e^(5x)e^y with respect to x is e^(5x)(5)*e^y, or 25x*e^y.
The partial of v = e^(5x)e^y with respect to y is e^(5x)e^y.
Answer: Brand B is
Explanation
(Brand A)24 diapers is 0.29¢ per diaper
(Brand B) 50 diapers is 0.27¢ per diaper
Answer:
y=7*3^x
Step-by-step explanation:
y=ab^x
(0,7) ⇒ 7=ab^0 ⇒ 7=a
(5, 1701) ⇒ 1701= 7b^5 ⇒ b^5=243 ⇒ b^5= 3^5 ⇒ b=3
y=7*3^x