The coordinates of point K(7, 0).
Given, JK has midpoint M(7, 2).
And, the coordinates of point J(7, 4).
We have to find the coordinates of point K.
As M is the midpoint, therefore by using the midpoint formula.
(xₙ + yₙ) = (x₁ + x₂/2 , y₁ + y₂/2)
Using x-coordinates,
xₙ = x₁ + x₂/2
7 = x₁ + 7/2
14 = x₁ + 7
or x₁ = 7
Nos using y-coordinates,
yₙ = y₁ + y₂/2
2 = y₁ + 4/2
4 = y₁ + 4
y₁ = 0
Therefore, the coordinates of point K(7, 0).
The coordinates of J(7, 4); K(7, 0); and M(7, 2).
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<span>h > t, u < t
can be written together either of these two ways:
u < t < h
or
h > t > u
And it could be any of these 120 numbers:
210 310 320 321 410 420 421 430 431 432 510 520
521 530 531 532 540 541 542 543 610 620 621 630
631 632 640 641 642 643 650 651 652 653 654 710
720 721 730 731 732 740 741 742 743 750 751 752
753 754 760 761 762 763 764 765 810 820 821 830
831 832 840 841 842 843 850 851 852 853 854 860
861 862 863 864 865 870 871 872 873 874 875 876
910 920 921 930 931 932 940 941 942 943 950 951
952 953 954 960 961 962 963 964 965 970 971 972
973 974 975 976 980 981 982 983 984 985 986 987
</span>
Answer:
its the 4th one . The one all the way to the bottom
Step-by-step explanation:
Answer: 
Step-by-step explanation:
Logarithms are in the format 
while exponential form is in the format 
So we can transfer a,b, and c from the log form to the exponential form
speed of current is 1.5 mi/hr
Answer:
let the rate in still water be x and rate of the current be y.
speed down the river is:
speed=distance/time
speed=14/2=7 mi/h
speed up the river is:
speed=(14)/(3.5)=4 mi/hr
thus total speed downstream and upstream will be:
x+y=7...i
x-y=4.......ii
adding the above equations i and ii we get:
2x=11
x=5.5 mi/hr
thus
y=5-5.5=1.5 mi/r
thus the speed in still waters is 5.5 mi/hr
speed of current is 1.5 mi/hr
