Pythagorean thoerem
we imagine the wall as a leg of a right triangle, the ladder is the hypotonuse and the distance from the bottom of the ladder to the wall is the other leg
so if we have a right triangle with leg length a and b and hypotnuse length c then
a²+b²=c²
so
given that one of the legs is 9ft (9 ft is the height of the triangle)
and the hypotnuse is 12 ft (length of ladder)
then say the last leg is x
so
x²+9²=12²
x²+81=144
mius 81 both sides
x²=63
sqrt both sides
x=√63
x=3√7
you should place it 3√7ft or about 7.9 or 8ft away
You say that there is uniform acceleration so:
vf-vi=at (final velocity minus initial velocity is equal to acceleration times time)
We know vf, vi, and t so we can solve for acceleration:
24-12=a10
12=10a
a=1.2
That is the acceleration, we will need to integrate with respect to time twice...
v=⌠a dt
v=at+vi , we know a=1.2m/s^2 and vi=12m/s
v=1.2t+12,
x=⌠1.2t+12 dt
x=1.2t^2/2+12t+xo, we can just let xo=0 for this problem...
x(t)=0.6t^2+12t
Now we know that this acceleration lasts for 10 seconds so the distance traveled in that time is:
x(10)=0.6(10^2)+12(10)
x(10)=60+120
x(10)=180 meters
Answer:
We are effectively looking for a and b such that 5, a, b, 135 is a geometric sequence.
This sequence has common ratio <span><span>3<span>√<span>1355</span></span></span>=3</span>, hence <span>a=15</span> and <span>b=45</span>
Explanation:
In a geometric sequence, each intermediate term is the geometric mean of the term before it and the term after it.
So we want to find a and b such that 5, a, b, 135 is a geometric sequence.
If the common ratio is r then:
<span><span>a=5r</span><span>b=ar=5<span>r2</span></span><span>135=br=5<span>r3</span></span></span>
Hence <span><span>r3</span>=<span>1355</span>=27</span>, so <span>r=<span>3<span>√27</span></span>=3</span>
Then <span>a=5r=15</span> and <span>b=ar=15⋅3=45</span>
Answer:
Step-by-step explanation:
Let the two consecutive numbers be 2n and 2n+2
The sum of these numbers in 2n + (2n+2).
Hence
2n + (2n+2) = 34
4n + 2 = 34
4n = 32
n =8.
If each of the 28 students made at least $25, you would multiply 28 and 25 together to obtain the least amount of money the class raised. That gets, 28x25 = 700. The class made at least $700.
