In the 
plane, we have 
everywhere. So in the equation of the sphere, we have
which is a circle centered at (2, -10, 0) of radius 4.
In the 
plane, we have 
, which gives
But any squared real quantity is positive, so there is no intersection between the sphere and this plane.
In the 
plane, 
, so
which is a circle centered at (0, -10, 3) of radius 
.
Answer:
Step-by-step explanation:
The distance from Q to S is 2 - (-14), or 16.
We start at point Q. Note how 3 and 5 add up to 8, which allows us to write:
R = Q + (3/8)(16), or R = -14 + 6, or R = -8.
From R to S it is (5/8)(16), or 10 units.
The directed line segment is partitioned into segments of lengths 6 and 10, whose combined length is 16, as expected.
Answer:
the 30th term is 239
Step-by-step explanation:
The computation of the 30th term is as follows:
As we know that
a_n = a_1 + (n-1)d
where
a_1 is the first number is the sequence
n = the term
And, d = common difference
Now based on this, the 30th term is
= 152 + (30 - 1) × 3
= 152 + 29 × 3
= 152 + 87
= 239
Hence, the 30th term is 239
20 √2
It would be a 45 45 90 triangle. So both sides are X (20) and the missing side is X √2 (20 √2).
Answer:
-7/9 or -.77777777....
Step-by-step explanation:
Just use keep change change
keep -5/9 the same
change the - to a +
change the 2/9 to a -2/9
5+2 = 7
put over 9
make negative
