Answer:
All points on line CD are equidistant from A and B
Step-by-step explanation:
Given that point A is the center of circle A and point B is the center of circle B, and the circumference of circle A passes through the center of circle B which is point B and vice versa.
Therefore we have;
The radius of circle A = The radius of circle B
Which gives;
The distance of the point C to the center A is equal to the distance of the point C to the center B
Similarly, the distance of the point D to the center A is equal to the distance of the point D to the center B
So also the distances of all points on the line from the center A is equal to the distances of all points on the line from the center B.
Answer:
Jacks scores 130 points
Jill scores 75 points
Step-by-step explanation:
J=Jacks score
K=Jills score
2K-20+K=205
So, we can do 3K-20=205
+20 +20
(cancel it out)
and we are left with 3K=225 which is K=75, so we do do 205-75=130(which is jacks score). To prove this we can do 75x2=150-20=130
Steve goes to a store to buy school supplies he buys 15 pencils the cost per pencil is .20 cents. How much did Steve spend to buy pencils
Answer:
2+8=10
Step-by-step explanation:
Answer:
Step-by-step explanation:
How did you determine that? One way is to carry out the division. However, that's laborious, especially with larger numbers. A better method is to look at the denominator of the simplified fraction. From here on, we will only be working with simplified fractions of the form
a
b
, where
gcd
(
a
,
b
)
=
1
and
b
>
1
. A well known fact is that if a prime
p
|
b
where
p
∉
{
2
,
5
}
, then the fraction is repeating. Think about why this is.
Here's a quick proof. If some fraction
a
b
has a terminating decimal representation, then there exists a positive integer
k
such that
10
k
⋅
a
b
=
n
, where
n
is a positive integer. Let
p
be a prime factor of
b
, and rewrite this as
10
k
a
=
n
b
. Then
p
|
10
k
a
, but since
gcd
(
a
,
b
)
=
1
,
p
∤
a
, so
p
|
10
k
(ie:
p
=
2
or
5
).
We will now proceed to the main part of this lesson: repeating decimals. We will define the period of a decimal
0.
d
1
d
2
d
3
d
4
…
to be the least positive integer
n
such that for some integer
k
,
d
i
=
d
i
+
n
for all
i
≥
k
. For example, the period of
1
90
=
0.0
¯¯¯
1
is
1
because for all
i
≥
2
,
d
i
=
d
i
+
1
=
1
. Similarly,
1
7
=
0.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
142857
has a period of
6
because
d
i
=
d
i
+
6
for all
i
≥
1
. An intuitive way of thinking about period is that it is the number of digits in the repetend when written optimally. ("Written optimally" informally means that you should write
0.0
¯¯¯
1
instead of
0.0
¯¯¯¯¯¯
11
.)
