Answer:
A
Step-by-step explanation:
i think?
Answer:
896
Step-by-step explanation:
Let's talk first about how many 3 digit numbers there are. The first 3 digit number is 100 and the last is 999. So there are 999-100+1 numbers that are 3 digits long. That simplifies to 900.
Now let's find how many of those have a sum for the digits being 1, then 2 ? Then take that sum away from the 900 to see how many 3 digit numbers have the sum of their digits being more than 2.
3 digit numbers with sum of 1:
The first and only number is 100 since 1+0+0=1.
We can't include 010 or 001 because these aren't really three digits long.
3 digit numbers with sum of 2:
The first number is 101 since 1+0+1=2.
The second number is 110 since 1+1+0=2.
The third number is 200 since 2+0+0=2.
That's the last of those. We could only use 0,1, and 2 here.... Anything with a 3 in it would give us something larger than or equal to 3.
So there are 900-1-3 numbers who are 3 digits long and whose sum of digits is greater than 2.
This answer simplifies to 896.
Answer:
i. RP = 17
ii. RT = 15
iii. RS = 30
Step-by-step explanation:
Given that; QT = 9 and TP = 8.
From the diagram, join R to P. Thus RP = QP (radius of the circle)
RP = QT + TP
= 9 + 8
RP = 17
Applying Pythagoras theorem to triangle TRP;
RT = 
= 
= 
= 15
∴ RT = 15
But, RT = TS = 15.
So that;
RS = 15 + 15
= 30
RS = 30
Therefore; RP = 17, RT = 15 and RS = 30.
Answer:
z=98 and 5x+57 and z are supplementary.
Step-by-step explanation:
vertical angles theorem
