Let L be the low copper alloy and H be the high copper alloy.
We need 0.15L + 0.60H = 0.42 and L + H = 100
L=100 - H
Substituting this into the second equation for L, we get:
0.15(100-H) + 0.60H = 0.42(H+L)
15 - 0.15H + 0.60H = 0.42(H+100-H)
15 + 0.45H = 0.42(100)
0.45H = 42 - 15 = 27
H = 27/.45 = 60 lbs
Back substitution, L = 40 lbs.
Answer:
choice 3) 50 in
Step-by-step explanation:
A = 625(Pi) in^2
radius = (sq rt 625)(Pi)/Pi
radius = 25
d = 2(radius)
d = 50 in
The answer is Choice A: (cos(theta), sin(theta))
More specifically x = cos(theta) and y = sin(theta) make up the point (x,y) which is where point P is located. Because we have a unit circle, the equation of the circle is x^2+y^2 = 1. This leads to cos^2(theta) + sin^2(theta) = 1 which is the pythagorean trig identity.
The percentage of time spend at least 80 minutes on homework will be 66.66%.
<h3>What is the percentage?</h3>
The Percentage is defined as representing any number with respect to the 100. It is denoted by the sign %.
Given that:-
The time he spent on homework each night. this box plot shows the results.
The time spent on homework (min.) 20 40 60 80 100 120.
The percentage will be calculated as:-
Maximum time spent = 120 minutes
The percentage for 80 minutes of homework will be:-
= 120 / 80
= 0.66=66 = 66.66%
Therefore the percentage of time spend at least 80 minutes on homework will be 66.66%.
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9x^3 + 30x^2 - 25x
step one: find the greatest common factor (GFC)
GFC=x
step two: factor out the GFC (write the GFC first. then in parentheses, divide each term by the GFC)
x(9x^3 over x + 30x^2 over x + -25x over x)
step three: simplify each term in parentheses
x(9x^2+30x-25) ...should be the answer
