Answer:
24 Domain: s>=2 or s<=-2
25. 3x^2 +14x +10
26. x^2 -2x+5
Step-by-step explanation:
24. Domain is the input or s values
square roots must be greater than or equal to zero
s^2-4 >=0
Add 4 to each side
s^2 >=4
Take the square root
s>=2 or s<=-2
25. f(g(x)) stick g(x) into f(u) every place you see a u
f(u) = 3u^2 +2u-6
g(x) = x+2
f(g(x) = 3(x+2)^2 +2(x+2) -6
Foil the squared term
= 3(x^2 +4x+4) +2x+4-6
Distribute
= 3x^2 +12x+12 +2x+4-6
Combine like terms
=3x^2 +14x +10
26 f(g(x)) stick g(x) into f(u) every place you see a u
f(u) = u^2+4
g(x) = x-1
f(g(x) = (x-1)^2 +4
Foil the squared term
= (x^2 -2x+1) +4
= x^2 -2x+5
Answer:
Point S is your answer.
Step-by-step explanation:
See how all of them are rays, but they all connect at point S.
Point S is your answer.
Answer:
0
Step-by-step explanation:
Let us take sin(60 degree + theta) =x
Substituting x in equation
=> x-x
=> 0
The maximum speed of a boat at 30 feet length of water is 0.093 nautical miles/hour or knots.
<u>Step-by-step explanation:</u>
- The equation for the maximum speed, s is given by s²= (16/9)x
- where, x is the length of the water line in feet.
It is given that, the modeled equation s²= (16/9)x is used to find the maximum speed in knots or nautical miles per hour.
The question is asked to find the maximum speed when the length of the water is 30 feet.
Therefore, to find the maximum speed in 30 feet water, the given modeled equation is used. So, substitute the 30 feet in place of x.
<u>Now, calculating the maximum speed :</u>
s² = (16/9)(30)
s² = 480 / 9
s² = 53.3
Taking square root on both sides,
s = √53.3
s = 7.3
The maximum speed of a boat at 30 feet length of water is 7.3 nautical miles/hour or knots.
Answer:
Number 4
Step-by-step explanation:
Side-Angle-Side, it has to be between the two sides.
