Answer:
the pay one
Step-by-step explanation:
i thinks sorry if i get it wrong
<h2><em>we can write (3x^2-5y^2) as (3x-5y)^2</em></h2><h2><em>(
3
x
−
5
y
)
2 as (
3
x−
5
y
)
(
3
x−
5
y
)</em></h2><h2><em>3
x
(
3
x
−
5
y
)
−
5
y
(
3x
−5
y
)</em></h2><h2><em>3
x
(
3
x
−
5
y
)
−
5
y
(3
x
−
5
y
)</em></h2><h2><em>3
x
(
3
x
)
+
3
x
(
−
5y
)
−
5
y
(
3
x
)
−
5
y(
-5
y
)</em></h2><h2><em>9
x
2
−
15
x
y
−
15y
x
+
25
y
2
</em></h2><h2><em> Subtract 15
y
x from −
15
x
y
.</em></h2><h2><em>9
x
2
−
30
xy
+
25
y
2</em></h2><h2><em> HOPE IT HELPS(◕‿◕✿) </em></h2><h2><em> SMILE!! </em></h2>
Given: p=2, h=8
5p^2=h
5(2)^2=8 ------- substitution here
5(4)=8
20 doesn't equal 8
I did exponents first because it comes before multiplication in the order of opertions.
Factor the following:x^4 + 4 x^3 + 6 x^2 + 4 x + 1
The coefficients match the 5^th row of Pascal's triangle, so x^4 + 4 x^3 + 6 x^2 + 4 x + 1 = (x + 1)^4:Answer: (x + 1)^4
