The value of x is 5, if a, b, and c are collinear points.
According to the given question.
a, b, and c are collinear points.
Which means points a, b, and c both lie in a same line.
b is between a nd c .
Also, ab = 12
bc = 5x - 2
and ac = 3x + 20
Since, b lies in between a and c
Therefore,
ab + bc = ac
12 + 5x - 2 = 3x + 20
⇒ 10 + 5x = 3x + 20
⇒ 5x - 3x = 20 - 10
⇒ 2x = 10
⇒ x = 10/ 2
⇒ x = 5
Therefore, the value of x is 5.
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Answer:
the first one. with two shaded dots and a line between them
Answer:
Step-by-step explanation:
In a G.P, the nth term is given as
Un=ar^(n-1)
Where
a is first term
n is nth term
And r is common ratio
So in the question given above,
The third term exceed the first term by 16
i.e U3=U1+16
Where U1=a. First term
U3=a+16
Given also that, the sum if the third term and fourth term is 72.
Then U3+U4=72.
We are told to find common ratio (r)
U3=ar^3-1
U3=ar^2
Also, U4=ar^3
U3+U4=72
ar^2+ar^3=72
ar^2(1+r)=72. equation 1
Also for
U3=a+16
ar^2=a+16
ar^2-a=16
a(r^2-1)=16. From (x^2-y^2)=(x+y)(x-y)
Then,
a(r-1)(r+1)=16. Equation 2
Divide equation 2 by equation 1
a(r-1)(r+1)/ar^2(1+r) =16/72
Then a cancel a and (1+r) cancel (1+r)
So,
(r-1)/r^2=2/9
Cross multiply
9(r-1)=2r^2
2r^2-9r+9=0
Solving the quadratic equation
2r^2-6r-3r+9=0
2r(r-3)-3(r-3)=0
(r-3)(2r-3)=0
r-3=0. Or. 2r-3= 0
Then r=3 or r=3/2
<span>JP + PK = JK
3y + 1 + 12y - 4 = 75
15y -3 = 75
15y = 75 +3
15y = 78
y = 78/15
y = 5.2
answer is </span><span>a. 5.2</span>