Answer:
1 is neither prime nor a composite number because it only has 1 divisor, 1 and 1, and it doesn't have more than 2 integral divisors, like stated, it only has 1. So it falls in neither category.
1)3(k+5)=-2(3k-6)
3k+15=-6k+12
3k=-6k+12-15
3k=-6k-3
3k+6k=-3
9k=-3
k=-3/9=-1/3
so, k=-1/3
2)1/2r+4=3/4r-3/2
1/2r=3/4r-3/2-4
1/2r=3/4r-11/2
1/2r-3/4r=-11/2
-r/4=-11/2
4*(-r/4)=4*(-11/2)
-r=-22
r=22
3)2a-9=2a+5
2a=2a+5+9
2a=2a+14
2a-2a=14
0=14
no solution
4)8m+2+4m=2(6m+1)
12m+2=12m+2
12m=12m+2-2
12m=12m
12m-12m=0
0=0
true for all m
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>
Answer:
A = 5,752.60
Step-by-step explanation:
P = $3,500
r= 5%
t = 10
k = 4
A = P(1 + r/k)^kt
= 3,500(1 + 0.05/4)^4*10
= 3,500(1 + 0.0125)^40
= 3,500(1.0125)^40
= 3,500(1.6436)
= 5,752.6
A = 5,752.60
