We want to find a point on a segment such that the point divides the segment in the ratio 5:1
We will find that the point T is (6, 9)
The general formula for the point T that divides a segment whose endpoints are (x₁, y₁) and (x₂, y₂) such that separates the segment in the ratio j:k is given by:
Here we have the endpoints (5, 10) and (11, 4), and we know that the ratio is 5:1
Then we can just use the above formula to get:
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Answer:
The answer is A
Step-by-step explanation:
You should set "I" as
1 0 0
0 1 0
0 0 1
then replace "a" to 0.5 and "b" to 0.1
then finally you multiply matrice C and D
If your answer is
1 0 0
0 1 0
0 0 1
then the answer will be A
Ho ho ho, lets get this party started
ok so I'm just really excited to use this stuff that I just learned
so
multiplicites
if a root or zero has an even multilicity, the graph bounces on that root
if the root or zero has an odd multiplicty, the graph goes through that root
so
roots are
-1
2
4
multiplicty is how many times it repeats
2 has even multiplity
we just do 2 is odd and 1 is even so
for roots, r1 and r2, the facotrs would be
(x-r1)(x-r2)
so
(x-(-1))^1(x-2)^2(x-4)
(x+1)(x-2)^2(x-4)
this is a 4th degre equaton
normally, it is goig from top right to top left
it is upside down
theefor it has negative leading coefient
y=-k(x+1)(x-4)(x-2)^2
Answer:
A.equal to
Step-by-step explanation:
thats the answer correct me if you want
Answer:
5/6
Step-by-step explanation: