2b+h = 4.82
b+2h = 3.70
multiply the second equation by 2:
2b + 4h = 7.40
subtract the first equation from it:
3h = 2.58
h = 0.86
Assuming T, B and N are all on the same line, then we can say
BT + TN = BN
which is the segment addition postulate.
Subtract TN from both sides to get
BT + TN = BN
BT + TN - TN = BN - TN
BT = BN - TN
BN - TN = BT
Which is what choice C is saying. Therefore the answer is choice C.
Answer:
C. 178−−−√ m
Step-by-step explanation:
Given the following :
v = final velocity (in m/s)
u = initial velocity (in m/s)
a = acceleration (in m/s²)
s = distance (in meters).
Find v when u is 8 m/s, a is 3 m/s², and s is 19 meters
Using the 3rd equation of motion :
v^2 = u^2 + 2as
v^2 = 8^2 + 2(3)(19)
v^2 = 64 + 114
v^2 = 178
Take the square root of both sides :
√v^2 = √178
v = √178
Answer:
88 problems
Step-by-step explanation:
Set this problem up as fractions
86/x = 98/100
Cross multiply
86 × 100 = x × 98
8600 = 98x
Divide both sides by 98
87.75... = x
Round to nearest whole number
87.75 rounds up to 88
88 problems
Answer:
2). As x-> -∞, f(x)->∞
As x-> ∞, f(x)-> -∞
5). As x-> -∞, f(x)-> -∞
As x-> ∞, f(x)-> ∞
3). As x-> -∞, f(x)-> -∞
As x-> ∞, f(x)-> ∞
6). As x-> -∞, f(x)-> ∞
As x-> ∞, f(x)-> ∞
Step-by-step explanation:
I just watched a quick video so you can't completely trust me, but i tried my best. Hopefully someone more trustworthy for this comes in.
