1. 8c^2-26c+15= (4c-3) (2c-5). Break the expression into groups: =(8c^2-6c)+(-20c+15). Factor out 8c^2-6c: 2c(4c-3). Factor out -5 from -20c+ 15: -5(4c-3). Lastly factor out common term (4c-3) and thats how you'll get your answer (4c-3) (2c-5).
2. common factors for 270 and 360 is 90.To find this write the factors of each and find the largest one.270: 1, 270, 2, 135, 3, 90, 5, 54, 6, 45, 9, 30, 10, 27, 15, 18360: 1, 360, 2, 180, 3, 120, 4, 90, 5, 72, 6, 60, 8, 45, 9, 40, 10, 36, 12, 30, 15, 24, 18, 20
3. The factors for 8 a3b2 and 12 ab4 is 4. because 8: 1, 2, 4, 812: 1, 2, 3, 4, 6, 12.
4. 81a^2+36a+4= (9a+2)^2. Break down the expression into groups: (81a^2+18a)+(18a+4). Factor out 9a from 81a^2 +18a: 9a(9a+2). Factor out 2 from 18a+4: 2(9a+2). so the groups you got are now 9a(9a+2)+2(9a+2). Lastley factor out common term (9a+2) to get (9a+2) (9a+2). Finally you get the answer (9a+2)^2.
5. mn-15+3m-5n= (n+3)(m-5). factor out m from nm+3m: m(n+3). Factor out -5 from -5n-15: -5(n+3). And thats how you get the number (n+3)(m-5)
Hope this helped :) Have a great day
Answer:
No, a triangle cannot be constructed with sides of 2 in., 3 in., and 6 in.
For three line segments to be able to form any triangle you must be able to take any two sides, add their length and this sum be greater than the remaining side.
2
in.
+
3
in.
=
5
in.
5
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6
in.
For a triangle with sides 3 in., 4 in. and 5 in. which can form a triangle:
3 + 4 = 7 which is greater than 5
3 + 5 = 8 which is greater than 4
4 + 5 = 9 which is greater than 3
Step-by-step explanation:
X = 8.5 because 8.5 x 2 = 17 and 17 + 1 = 18
Answer:
its simple its C
Step-by-step explanation:
forget humanity and return to monke
Answer:
264
Step-by-step explanation:
the perimeter of a square with sides of 61+5 is 264