A negative plus a negative equals a negative.
A negative plus a positive depends on which number is bigger for the sign.
A negative times a negative equals a positive.
A negative times a positive equals a negative.
A negative divided by a negative equals a positive.
A negative divided by a positive equals a negative.
A negative minus a negative depends on which number is bigger for the sign.
A negative minus a positive equals a negative.
Solution
To find Which two numbers on the number line has an
absolute value of 3
Absolute value means " how far a number is from
zero " .
"3" is 3 away from zero .
and " -3 " is also 3 away from zero .
i.e
| 0 - 3 | = 3
| 3 - 0 | = 3
So , the absolute value of "3 " is 3
similarly , absolute value of "3 " is 3 .
Hence , the two numbers are -3 and 3
Answer:
132 i think i hope this helps
Step-by-step explanation:
3*12=36
36/2=18
12*8=96
96+36=132
Perimeter (p) = 2×length (l) + 2×width (w)
p = 2l+2w
area (a) = l×w, so solve for one (I'll use l):
![a = l \times w \\ l = a \div w \\ p = 2l + 2w = 2(l + w)](https://tex.z-dn.net/?f=a%20%3D%20l%20%5Ctimes%20w%20%5C%5C%20l%20%3D%20%20a%20%5Cdiv%20w%20%5C%5C%20p%20%3D%202l%20%2B%202w%20%3D%202%28l%20%2B%20w%29)
since p = 30, and a = 50, substitute the "a÷w" in for l in the perimeter equation:
![p = 2(l + w) = 2((a \div w) + w) \\ = 2((a \div w) + ( {w}^{2} \div w)) \\ p= 2((a + {w}^{2}) \div w](https://tex.z-dn.net/?f=p%20%3D%202%28l%20%2B%20w%29%20%3D%202%28%28a%20%5Cdiv%20w%29%20%2B%20w%29%20%5C%5C%20%20%3D%202%28%28a%20%5Cdiv%20w%29%20%2B%20%28%20%7Bw%7D%5E%7B2%7D%20%5Cdiv%20w%29%29%20%5C%5C%20%20p%3D%202%28%28a%20%2B%20%20%7Bw%7D%5E%7B2%7D%29%20%5Cdiv%20w)
Now plug in p and a values:
![p= 2((a + {w}^{2}) \div w) \\ 30 = 2((50 + {w}^{2}) \div w) \\ 30 \div 2 = (50 + {w}^{2}) \div w \\ 15w = 50 + {w}^{2}](https://tex.z-dn.net/?f=p%3D%202%28%28a%20%2B%20%20%7Bw%7D%5E%7B2%7D%29%20%5Cdiv%20w%29%20%5C%5C%2030%20%3D%202%28%2850%20%2B%20%20%7Bw%7D%5E%7B2%7D%29%20%5Cdiv%20w%29%20%20%5C%5C%2030%20%5Cdiv%202%20%3D%20%2850%20%2B%20%20%7Bw%7D%5E%7B2%7D%29%20%5Cdiv%20w%20%5C%5C%2015w%20%3D%2050%20%2B%20%20%7Bw%7D%5E%7B2%7D%20%20)
![15w = 50 + {w}^{2} \\ {w}^{2} - 15w + 50 = 0 \\ (w - 5)(w - 10) = 0](https://tex.z-dn.net/?f=15w%20%3D%2050%20%2B%20%20%7Bw%7D%5E%7B2%7D%20%20%5C%5C%20%20%7Bw%7D%5E%7B2%7D%20%20-%2015w%20%2B%2050%20%3D%200%20%5C%5C%20%28w%20-%205%29%28w%20-%2010%29%20%3D%200)
therefore width can be either 5 or 10 (but not both), so let's plug in:
l = a÷w = 50÷5 = 10
So if w = 5, then l = 10
D) 5 feet, and 10 feet
Answer:
2,400 cm²
Step-by-step explanation:
You can solve for x because you know that 40cm=x²+4cm:
36 cm = x²
x = 6cm.
Therefore the width is 2(6)²-12, which is 60cm.
So the area is 40 cm x 60 cm, which is 2,400 cm².