The first 3 are examples of the difference of 2 squares so you use the identity
a^2 - b^2 = (a + b)(a - b)
x^2 - 49 = 0
so (x + 7)(x - 7) = 0
so either x + 7 = 0 or x - 7 = 0
giving x = -7 and 7.
Number 7 reduces to 3x^2 =12, x^2 = 4 so x = +/- 2
Number 8 take out GCf (d) to give
d(d - 2) = 0 so d = 0 , 2
9 and 10 are more difficult to factor
you use the 'ac' method Google it to get more details
2x^2 - 5x + 2
multiply first coefficient by the constant at the end
that is 2 * 2 = 4
Now we want 2 numbers which when multiplied give + 4 and when added give - 5:- -1 and -4 seem promising so we write the equation as:-
2x^2 - 4x - x + 2 = 0
now factor by grouping
2x(x - 2) - 1(x - 2) = 0
(x - 2) is common so
(2x - 1)(x - 2) = 0
and 2x - 1 = 0 or x - 2 = 0 and now you can find x.
The last example is solved in the same way.
Answer:
Step-by-step explanation:
AD = DC and BE=EC ⇒
DE = 1/2AB
- 4x + 1 = 1/2(11x - 25)
- 2(4x + 1) = 11x - 25
- 8x + 2 = 11x - 25
- 11x - 8x = 2 + 25
- 3x = 27
- x = 27/3
- x = 9
DE = 4*9 + 1 = 36 + 1 = 37
224 i think because we are in the same class and i like boogers and poop
Answer: 108
lmk if this is wrong
Step-by-step explanation:
V=1/3bh
V=1/3(1/2bh)(h)
V=1/3(1/2)(9)(6)(12)
(i simplified)
V=(1)(1)(3)(6)(6)
V=108
100-45+22= 77
you will have 77 percent of battery on your phone
