Answer:
<h2>d(4) = 64</h2>
Step-by-step explanation:
d(1) = 13
d(n) = d(n - 1) + 17
where
n is the nth term
To find the 4th term we must first find the second term use it to find the 3rd term and use the 3rd term to find the 4th term
So we have
<u>For the 2nd term</u>
That's d(2)
We have
d(2) = d(2-1) + 17
d(2) = d(1) + 17
But d(1) = 13
d(2) = 13 + 17 = 30
<u>For the 3rd term d(3)</u>
That's
d(3) = d(3 - 1) + 17
d(3) = d(2) + 17
d(3) = 30 + 17 = 47
<u>For the 4th term that's d(4)</u>
We have
d(4) = d(4 - 1) + 17
d(4) = d(3) + 17
d(4) = 47 + 17
We have the final answer as
<h3>d(4) = 64</h3>
Hope this helps you
Answer:
There are 15 men in the dance recital.
Step-by-step explanation:
If 20 people is 4 parts, then 5 people are 1 part. Due to that, 15 people will equal 3 parts.
Answer:
17.3 in
Step-by-step explanation:
Answer:
The absolute value of a number is the distance between the number and zero on a number line. In other words, a number and its opposite have the same absolute value.
Step-by-step explanation:
see the attached figure with the letters
1) find m(x) in the interval A,BA (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100
2) find m(x) in the interval B,CB(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20
3)
find n(x) in the interval A,BA (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x
4) find n(x) in the interval B,CB(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30
5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x)
</span>h'(x)=-36/25=-1.44
6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72
for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72
<span> h'(x) = 1.44 ------------ > not exist</span>