Answer:
Step-by-step explanation:
a.
first number is 1000-1+9=1008
9)1000(1
9
-------
10
9
-----
10
9
----
1
----
last number is 9999
9| 9999
---------
1111 |0
--------
9999=1008+(n-1)9
9999-1008=(n-1)9
n-1=8991/9=999
n=999+1=1000
b.
first digit=1000
last digit=9999-1=9998
2 |9999
---------
|4999|1
9998=1000+(n-1)2
(n-1)2=9998-1000=8998
n-1=4499
n=4499=1=5000
c.not sure
d.
total numbers=9000
9999=1000+(n-1)1
9999-1000=n-1
n=8999+1=9000
numbers divisible by 3=3000
first number=1002
last number=9999
9999=1002+(n-1)3
(n-1)3=9999-1002=8997
n-1=2999
n=2999+1=3000
numbers not divisible by 3=9000-3000=6000
e.
numbers divisible by 5=1800
first number=1000
last number=9995
9995=1000+(n-1)5
(n-1)5=9995-1000=8995
n-1=1799
n=1799+1=1800
numbers divisible by 7=1286
7 | 1000
---------
| 142-6
1000-6+7=1001
7 | 9999
|---------
1428-3
9999-3=9996
first digit=1001
last digit=9996
9996=1001+(n-1)7
(n-1)7=9996-1001=8995
n-1=1285
n=1285+1=1286
numbers divisible by 35=257
first digit=1015
35 ) 1000 ( 28
70
----
300
280
------
20
---
1000-20+35=1015
35)9999(285
70
----
299
280
-----
199
175
----
24
----
last digit=9999-24=9975
9975=1015+(n-1)35
(n-1)35=9975-1015=8960
n-1=8960/35=256
n=257
reqd. numbers=1800+1286-257=3019
Answer:
It will be $40.00
Step-by-step explanation:
hope this helps
Select all of the relations that are functions. {(a, 1), (b, 1), (c, 1)} {(a, a), (a, b), (a, c)} {(1, a), (2, a), (3, a)} {(a,
netineya [11]
Answer:
{(a, 1), (b, 1), (c, 1)}, {(1, a), (2, a), (3, a)}, and {(a, a), (b, b), (c, c)}
**All functions are relations, but not ll relations are functions**
In a function, the x values CAN NOT repeat, only y-values can.
48m^5n: 3*2^4*m^5n
81m^2n^2: 3^4m^2n^2
GCF: 3*m^2n
Your answer is 76m.
a^2+b^2=c^2
57^2+b^2=95^2
3249+b^2=9025
b^2=5776
Take sqr on both sides and you get:
b=76m
