Answer:
1. 6
2. 23
Step-by-step explanation:
1.
logx+1 ( x^2 + 2x + 1)^3
= 3logx+1 ( x^2 + 2x + 1)
= 3 logx+1 (x + 1)^2
= 6 logx+1 (x + 1) Now the logx+1 x+1 = 1 so
The answer is 6.
2.
log11 11^23
= 23 log11 11
As in question 1 log11 11 = 1
So the answer is 23.
Answer:
Step-by-step explanation:
The usual equation used for the vertical component of ballistic motion is ...
h(t) = -16t² +v₀t +h₀
where v₀ is the initial upward velocity, and h₀ is the initial height. Units of distance are feet, and units of time are seconds.
Your problem statement gives ...
v₀ = 64 ft/s
h₀ = 8 ft
so the equation of height is ...
h(t) = -16t² +64t +8
__
For quadratic ax² +bx +c, the axis of symmetry is x=-b/(2a). Then the axis of symmetry of the height equation is ...
t = -64/(2(-16)) = 2
The object will reach its maximum height after 2 seconds.
The height at that time will be ...
h(2) = -16(2²) +64(2) +8 = 72
The maximum height will be 72 feet.
Right triangle
so
n + n + 40 = 90
2n = 50
n = 25
answer
Measure of angle n = 25°
Answer:
The corresponding point is (1 , -8)
Step-by-step explanation:
∵ Point (4 , 8) on the graph of f(x)
∵ f(x+3) means the graph of f(x) moved horizontally 3 units to the left
∵ x-coordinate = 4 in f(x)
∴ x-coordinate = 4 - 3 = 1 in f(x+3)
∴The point is (1 , -8)
<span>The flagpole stands vertically on the ground. If a line is drawn from the top of the flagpole to the shadow, it forms a right-angled triangle.
Similarly, if a line is drawn from the top of the student's head to the tip of this shadow a right-angled triangle is formed.
Since it is given that the measures of both the shadows are taken at the same time and location, the angle of depression in both the cases is equal.
Therefore, by AA similarity,
height of the flagpole/ height of the student =length of shadow of the flagpole/length of shadow of the student
flagpole height/1.6m = 8.5 m/2.0m
flagpole height = (8.5 m x 1.6m )/2.0 m
=6.8m</span>
