We are given
arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103
so, first term is 124
u(1)= 124
now, we can find common difference
now, we can find kth term
now, we can plug values
and we get
u(k) must be negative
so,
now, we can solve for k
so, it's closest integer value is
..............Answer
For this case we have that the original point is given by:
B = (7, 2)
As the point is reflected through the x axis, then we have the following transformation:
(x, y) --------------> (x, -y) -------------> (x ', y')
Applying the transformation to the original ordered pair we have:
(7, 2) --------------> (7, - (2)) -------------> (7, -2)
Answer:
Point B 'is given by:
B '= (7, -2)
1) 2(6-8x)
2) 4(3-4x)
That seems about right.
Step-by-step explanation:
JNM is a right-angled triangle with JM being the Hypotenuse.
we know JN = 20
NM = KL, and that is a side of the right-angled triangle JKL, aber we know JK and JL.
KL² = JL² - JK² = 35² - 16² = 969
KL = NM = sqrt(969) = 31.12876483...
and now,
JN² + NM² = JM²
20² + 969 = JM²
400 + 969 = JM²
1369 = JM²
JM = 37
