Factor?, if so
if in ax^2+bx+c form then
multiply a and c together
result=z,
find 2 numbers that multiply to get z and add up to get b
so
14=a
-17=b
-6=c
14 times -6=-84
factor to find what 2 numbers add up to -17
-84=
-2, 42
-3,28
-4,21
-6,14
-7,12
add each up
-2, 42=40 nope
-3,28=25 nope
-4,21=17 yes, becase if -21 and 4 then answer
-6,14=8 nope
-7,12=5 nope
split up -17 into -21 and 4
14x^2-21x+4x-6
group
(14x^2-21x)+(4x-6)
undistribute
(7x)(2x-3)+(2)(2x-3)
revesres undistribute (ab+ac=a(b+c) or (a+b)(c)+(a+b)(d)=(a+b)(c+d))
(7x+2)(3x-3) is factore out form
then set it to zero and do th emath because if xy=0 then assume x and/or y=0 so
7x+2=0
subtract 2
7x=-2
divide by 7
x=-2/7
3x-3=0
add 3
3x=3
divide 3
x=1
x=-2/7 or 1
Answer:
There will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days
Step-by-step explanation:
The given parameters are;
The half life of the radioactive substance = 45 days
The mass of the substance initially present = 6.2 grams
The expression for evaluating the half life is given as follows;

Where;
N(t) = The amount of the substance left after a given time period = 1 gram
N₀ = The initial amount of the radioactive substance = 6.2 grams
= The half life of the radioactive substance = 45 days
Substituting the values gives;




The time that it takes for the mass of the radioactive substance to remain 1 g ≈ 118.45 days
Therefore, there will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days.
Answer:
2561
Step-by-step explanation:
3+5=8
8 to the 3rd power= 512
512 times 5= 2560
2560+1= 2561
It's (5,2) , I took it already so here you go
Answer:
jksahfoahsofahsdubouahdkj
Step-by-step explanation: