We can only solve it is equal to something
if we assume that it eausl zero, then if we factor it we can assume the factors eqal zero
xy=0, then assume x and y=0
so what we do is if we have
ax^2+bx+c
we factor a and take the factors closest together
then place them with the x
then factor c and chose the closest ones
then try them and see if it works
so if we had
6x^2+20x+16
6=2 times 3
(2x+ )(3x+ )
we know that they are both + because the middle terms is positive and the end terms is positive
16=4 times 4
(2x+4)(3x+4)=6x^2+12x+8x+16=6x^2+20x+16
th easy way to check is we know that teh end terms will be right since we made them that way
to check the middle terms
if we have
(ax+b)(cx+d)
to check what the middle term is we do
bcx+dax
and see if that equals the middle term
so
2. 6x^2+20x+16
6=2 times 3
(2x+ )(3x+ )
we know that they are both + because the middle terms is positive and the end terms is positive
16=4 times 4
(2x+4)(3x+4)=6x^2+12x+8x+16=6x^2+20x+16
facored is (2x+4)(3x+4)=0
2x+4=0
x=-2
3x+4=0
x=-4/3
3. 5x^2+21x+4
5=1 times 5 prime is nice since we only have 1 option
(x+ )(5x+ )
we know they are both pluss since middle and end term is positive
4=2 times 2
(x+2)(5x+2)
(2 times 5x)+(x times 2)=10x+2x=12x, we need 21 so this wrong
change end term
(x+4)(5x+1)
(4 times 5x)+(1 times x)=20x+x=21x correct
factore form
(x+4)(5x+1)=0
x+4=0
x=-4
5x+1=0
x=-1/5
4.
-7x^2+59x-24
-7=-1 and 7 or -7 and 1
we pick -1 and 7 since we want the middle term to be positive
(7x- )(-x+)
signs are differenc since end term is negaive
-24=8 and -3
(7x-3)(-x+8)
(7x times 8)+(-3 times -x)=56x+3x=59x
correct
(7x-3)(-x+8)=0
7x-3=0
x=3/7
-x+8=0
x=8
OK, so I am getting tired of this so I will just do the rest in factored form isnce you probably shoud have figured out how to do it from the previous examples
5. (x-3)(3x-5)
6. (x-1)(2x+13)
7. (x-2)(3x+8)
8. 2x^2+17x-20 cannot be factored but 2x^2+18x-20 can maybe typing mistake
9. (x+3)(11x-8)
answers:
2.(2x+4)(3x+4)
3.(x+4)(5x+1)
4.(7x-3)(-x+8)
5. (x-3)(3x-5)
6. (x-1)(2x+13)
7. (x-2)(3x+8)
8. 2x^2+17x-20 cannot be factored but 2x^2+18x-20 can maybe typing mistake
9. (x+3)(11x-8)
Answer: (B) 130 cm
Proof: If a chord and a line from the centre meet, they must be perpendicular. The line from the centre to the chord also bisects the chord. If the distance from the centre of the circle to the chord is 52 cm and the distance of half of the chord is 39 cm, then you can work out using Pythagoras' Theorem the distance of the radius.
r² = 52² + 39²
r² = 4225
r = 65
Thus, the diameter is 2r = 130 cm
4 x > − 16 (Possibility 1)
4 x
4 > − 16
4 (Divide both sides by 4)
x > − 4
6 x ≤ − 48 (Possibility 2)
6 x
6 ≤ − 48
6 (Divide both sides by 6)
x ≤ − 8
Answer:
x > − 4 or x ≤ − 8
Answer:
The supplementary angle would be 40°.
Step-by-step explanation:
Supplementary angles equal 180°
180 - 140
40
Answer:quick it’s 210 (:
Step-by-step explanation:
Your welcome ask more questions if you need help good luck btw u double checked with an app thank me later