<h3>I'll teach you how to solve 15x ^2 −4x−4</h3>
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15x ^2 −4x−4
Break the expressions into groups:
(15x^2+6x) + (-10x-4)
Factor out 3x from 15x^2+6x:
3x(5x+2)
Factor out -2 from -10x-4:
-2(5x+2)
3x(5x+2) -2(5x+2)
Factor out common term 5x+2:
(5x+2)(3x-2)
Your Answer Is (5x+2)(3x-2)
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To get the time taken for Jan and Jo to meet we proceed as follows;
Jan's speed=70 mph
Jo's speed =60 mph
distance between =1170
relative speed=70+60=130
time taken for them to meet will be:
time=(distance)/(relative speed)
=1170/130
=9 hours
the answer is 9 hours
To get started, we will use the general formula for bacteria growth/decay problems:

where:
A_{f} = Final amount
A_{i} = Initial amount
k = growth rate constant
t = time
For doubling problems, the general formula can be shortened to:

Now, we can use the shortened formula to calculate the growth rate constant of both bacteria:
Colby (1):


per hour
Jaquan (2):


per hour
Using Colby's rate constant, we can use the general formula to calculate for Colby's final amount after 1 day (24 hours).
Note: All units must be constant, so convert day to hours.


Remember that the final amount for both bacteria must be the same after 24 hours. Again, using the general formula, we can calculate the initial amount of bacteria that Jaquan needs:

5
3
5
−
3
5
+
−
6
7
−
(
−
1
)
=
28
5
−
3
5
+
−
6
7
−
(
−
1
)
=
5
+
−
6
7
−
(
−
1
)
=
29
7
−
(
−
1
)
=
36
7
=
5
1
7