A = a^2.....where " a " is the size of 1 side
A = 256
256 = a^2...take sqrt of both sides, eliminating the ^2
sqrt 256 = a
16 = a
so each side is 16 ft....now we find the perimeter
P = 4a....where " a " is the size of each side
P = 4(16)
P = 64.......so they will need 64 ft of fencing <===
Answer:
x=105
Step-by-step explanation:
3x+4y=15
3x+4y=7
7*15=105
x= 105
i really hopes this helps i did this in last semester
Answer:
The ratio of the first term to the common difference is 1 : 2
Step-by-step explanation:
The rule of the sum of the AP is
![S_{n}=\frac{n}{2}[2a+(n-1)d]](https://tex.z-dn.net/?f=S_%7Bn%7D%3D%5Cfrac%7Bn%7D%7B2%7D%5B2a%2B%28n-1%29d%5D)
, where
a is the first number
d is the common difference
n is the position of the number
∵ 
= 4 × 
⇒ (1)
→ Find and
∵ In n = 10
∴ ![S_{10}=\frac{10}{2}[2a+(10-1)d]](https://tex.z-dn.net/?f=S_%7B10%7D%3D%5Cfrac%7B10%7D%7B2%7D%5B2a%2B%2810-1%29d%5D)
∴ ![S_{10}=5[2a+9d]](https://tex.z-dn.net/?f=S_%7B10%7D%3D5%5B2a%2B9d%5D)
∴ = 10a + 45d ⇒ (2)
∵ In n = 5
∴ ![S_{5}=\frac{5}{2}[2a+(5-1)d]](https://tex.z-dn.net/?f=S_%7B5%7D%3D%5Cfrac%7B5%7D%7B2%7D%5B2a%2B%285-1%29d%5D)
∴ ![S_{5}=2.5[2a+4d]](https://tex.z-dn.net/?f=S_%7B5%7D%3D2.5%5B2a%2B4d%5D)
∴ = 5a + 10d ⇒ (3)
→ Substitute (2) and (3) in (1)
∵ 10a + 45d = 4[5a + 10d]
∴ 10a +45d = 20a + 40d
→ Subtract 40d from both sides
∴ 10a + 45d - 40d = 20a + 40d - 40d
∴ 10a + 5d = 20a
→ Subtract 10a from both sides
∴ 10a - 10a + 5d = 20a - 10a
∴ 5d = 10a
→ Divide both sides by 5
∴ d = 2a
→ That means d is twice a or a is half d
∴ a : d = 1 : 2
The ratio of the first term to the common difference is 1 : 2
Answer- since you are buying x packs of pencils, 2x packs of erasers, and 3x rolls of tape, you would be buying a total of 6x items.
D) 2g² + 5g + 4 is the polynomial that is factored completely.
This is because the terms in the polynomial do not have any common factors.
a) g⁵ - g can be factored into g(g⁴ - 1)
b) 4g³ + 18g² + 20g can be factored into 2g(2g² + 9g + 10)
c) 24g² - 6g⁴ cab be factored into 6g²(4-g²)
