We know that
Part a) <span>Find the fifth term of the arithmetic sequence in which t1 = 3 and tn = tn-1 + 4
t1=3
t2=t1+4----> 3+4-----> 7
t3=t2+4-----> 7+4----> 11
t4=t3+4-----> 11+4---> 15
t5=t4+4-----> 15+4---> 19
the answer Part a) is 19
Part b) </span><span>Find the tenth term of the arithmetic sequence in which t1 = 2 and t4 = -10
we know that
tn=t1+(n-1)*d-----> d=[tn-t1]/(n-1)
t1=2
t4=-10
n=4
find the value of d
d=[-10-2]/(4-1)-----> d=-12/3----> d=-4
find the </span>tenth term (t10)
t10=t1+(10-1)*(-4)----> t10=2+9*(-4)----> t10=-34
the answer Part b) is -34
Part c) <span>Find the fifth term of the geometric sequence in which t1 = 3 and tn = 2tn-1
t1=3
t2=2*t1----> 2*3----> 6
t3=2*t2----> 2*6----> 12
t4=2*t3-----> 2*12---> 24
t2=2*t4----> 2*24----> 48
the answer Part c) is 48</span>
11/12 hours left. 1 2/3 - 3/4 = 11/12. First, find the common denominator (12) . The problem is now 1 8/12 - 9/12. Use the 1 in (1 8/12) to borrow. Multiply the denominator by 1 (12) and add that to the numerator. The question is now 20/12 - 9/12. I assume that you can finish it off from there.
First, we get the length of the runway in which the pilot had seen a 45° by using the trigonometric formula of tangent.
tan 45° = 1,500 / x
The value of x from the equation is 1500 ft.
We do the same for the 42°
tan 42° = 1,500 / y
The value of y is equal to 1665.92 ft. To determine the length of the runway, we subtract the lengths calculated and this will give us an answer of 165.92 ft.
Answer:
7 4/17
Step-by-step explanation:
