Answer:
2.1
Step-by-step explanation:
By the Law of Cosines,
a² = 5² + 6² - 2(5)(6)cos(20°)
a = sqrt(5² + 6² - 2(5)(6)cos(20°)), which is about 2.1
A=(-1,4)=(xa,ya)→xa=-1, ya=4
B=(-2,1)=(xb,yb)→xb=-2, yb=1
C=(2,1)=(xc,yc)→xc=2, yc=1
Perimeter <span>of ∆ABC: P=AB+BC+AC
AB=d A-B=sqrt [ (xb-xa)^2+(yb-ya)^2 ]
AB=sqrt [ (-2-(-1))^2+(1-4)^2]
AB=sqrt [ (-2+1)^2+(-3)^2]
AB=sqrt [ (-1)^2+9]
AB=sqrt [ 1+9]
AB=sqrt [10]
AB=3.162277660
</span>BC=d B-C=sqrt [ (xc-xb)^2+(yc-yb)^2 ]
BC=sqrt [ (2-(-2))^2+(1-1)^2]
BC=sqrt [ (2+2)^2+(0)^2]
BC=sqrt [ (4)^2+0]
BC=sqrt [ 16+0]
BC=sqrt [16]
BC=4
AC=d A-C=sqrt [ (xc-xa)^2+(yc-ya)^2 ]
AC=sqrt [ (2-(-1))^2+(1-4)^2]
AC=sqrt [ (2+1)^2+(-3)^2]
AC=sqrt [ (3)^2+9]
AC=sqrt [ 9+9]
AC=sqrt [9*2]
AC=sqrt [9] * sqrt [2]
AC=3 sqrt [2]
AC=3 (1.414213562)
AC=4.242640686
P=AB+BC+AC
P=3.162277660+4+4.242640686
P=11.40491834
To the nearest tenth:
P=11.4
Answer: <span>The perimeter of ∆ABC is 11.4 units</span>
Answer:a
Step-by-step explanation:
Answer:
641 times
Step-by-step explanation:
Rate of transfer in 1989
2.4 kilobytes per second
Rate of transfer in a typical modem
1000 kilobyte per 0.65 seconds
Transfer per second for typical modem
1000/0.65=1538.461538 kilobyte per second
Comparing typical modem to 1989
=641.025641 times faster
Rounding off, this is 641 times faster
Therefore, ans=641 times
