Answer:
Reflection over the Y axis-
Step-by-step explanation:
Answer:
![Perimeter = 21](https://tex.z-dn.net/?f=Perimeter%20%3D%2021)
Step-by-step explanation:
Given
![IJ = 10](https://tex.z-dn.net/?f=IJ%20%3D%2010)
![FH = 8](https://tex.z-dn.net/?f=FH%20%3D%208)
![GH =14](https://tex.z-dn.net/?f=GH%20%3D14)
Required
Calculate the perimeter of IJH
<em>See attachment for illustration</em>
From the attachment,
I = mid of FH.
So:
![HI = \frac{1}{2} * FH](https://tex.z-dn.net/?f=HI%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20FH)
![HI = \frac{1}{2} * 8](https://tex.z-dn.net/?f=HI%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%208)
![HI = 4](https://tex.z-dn.net/?f=HI%20%3D%204)
J = mid of GH.
So:
![HJ = \frac{1}{2} * GH](https://tex.z-dn.net/?f=HJ%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20GH)
![HJ = \frac{1}{2} * 14](https://tex.z-dn.net/?f=HJ%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%2014)
![HJ = 7](https://tex.z-dn.net/?f=HJ%20%3D%207)
Perimeter IJH is then calculated as:
![Perimeter = HI + HJ + IJ](https://tex.z-dn.net/?f=Perimeter%20%3D%20HI%20%2B%20HJ%20%2B%20IJ)
![Perimeter = 4 + 7 + 10](https://tex.z-dn.net/?f=Perimeter%20%3D%204%20%2B%207%20%2B%2010)
![Perimeter = 21](https://tex.z-dn.net/?f=Perimeter%20%3D%2021)
The value of x is 9.
2*9 + 3*6 = 18 + 18 = 36.
<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>
Answer:
11a^3 + 7a^2 + 8a
Step-by-step explanation:
( 5a^3 - 2a^2 ) + ( 6a^3 + 9a^2 + 8a )
= 5a^3 + 6a^3 + -2a^2 + 9a^2 + 8a
= 11a^3 + 7a^2 + 8a