74% is greater than 11/15
O.6 easy because of the answer that is why
Answer:
Q13. y = sin(2x – π/2); y = - 2cos2x
Q14. y = 2sin2x -1; y = -2cos(2x – π/2) -1
Step-by-step explanation:
Question 13
(A) Sine function
y = a sin[b(x - h)] + k
y = a sin(bx - bh) + k; bh = phase shift
(1) Amp = 1; a = 1
(2) The graph is symmetrical about the x-axis. k = 0.
(3) Per = π. b = 2
(4) Phase shift = π/2.
2h =π/2
h = π/4
The equation is
y = sin[2(x – π/4)} or
y = sin(2x – π/2)
B. Cosine function
y = a cos[b(x - h)] + k
y = a cos(bx - bh) + k; bh = phase shift
(1) Amp = 1; a = 1
(2) The graph is symmetrical about the x-axis. k = 0.
(3) Per = π. b = 2
(4) Reflected across x-axis, y ⟶ -y
The equation is y = - 2cos2x
Question 14
(A) Sine function
(1) Amp = 2; a = 2
(2) Shifted down 1; k = -1
(3) Per = π; b = 2
(4) Phase shift = 0; h = 0
The equation is y = 2sin2x -1
(B) Cosine function
a = 2, b = -1; b = 2
Phase shift = π/2; h = π/4
The equation is
y = -2cos[2(x – π/4)] – 1 or
y = -2cos(2x – π/2) - 1
Answer:
The population density is 0.364 fish per gallon.
Step-by-step explanation:
The population density can be defined as the number of individuals (in this case, small tropical fish) per unit of area or volume. In this case, the volume of the aquarium (55 gallon).
The density can be calculated then as:
The population density is 0.364 fish per gallon.
Answer:
Step-by-step explanation:
26% is (26-20)/(50-20) = 6/30 = 1/5 of the way between 20% and 50%. That means 1/5 of the solution is 50% acid.
50% acid: 1/5 · 100 mL = 20 mL
20% acid: 100 mL -20 mL = 80 mL
Delbert must mix 80 mL of 20% acid and 20 mL of 50% acid.
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Maybe you'd like to see an equation. Let x represent the amount of 50% acid required. Then 100-x is the amount of 20% acid needed. The amount of acid in the mix is ...
0.50(x) +0.20(100 -x) = 0.26(100)
(0.50 -0.20)x = (0.26 -0.20)100 . . . . subtract 0.20(100)
x = (0.26 -0.20)/(0.50 -0.20)×100 = 20
This last expression should look a lot like the one we started with in this answer. It shows you how you can almost write down the answer to mixture problems without a lot of work.
