Answer:
see below
Step-by-step explanation:
amount of water w
l liter of pure alcohol = 100% alcohol
solution is 45 % percent alcohol
total amount of fluid is w+l
(w+l)( .45) = l*.100
Distribute
.45w + .45 l = 1l
.45 w = 1l - .45l
.45 w = .55l
w = .55l / .45
w =11/9 l
What one do you need help with?
Answer:
arcsin0.616
Step-by-step explanation:
arcsino.616
Not necessarily.
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and
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may be linearly dependent, so that their span forms a subspace of

that does not contain every vector in
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.
For example, we could have
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and

. Any vector
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of the form

, where

, is impossible to obtain as a linear combination of these
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and

, since

unless

and
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.
Answer:
Jose ends up with more money with $59 more than Peter.
Step-by-step explanation:
To determine the amount they will have, you have to use the formula to calculate the future value:
FV=PV(1+r)^n
FV= future value
PV= present value
r= rate of interest
n= number of periods of time
-Peter:
FV=1,000*(1+0.04)^10
FV=1,000*1.48
FV=1,480
-Jose:
FV=900*(1+0.05)^11
FV=900*1.71
FV=1,539
Difference: $1,539-$1,480=$59
According to this, the answer is that Jose ends up with more money with $59 more than Peter.