Answer:
30 unit
I hope I can help
Step-by-step explanation:
i hope I can help
I am not 100% positive but I think it is 4117.66

Let's find out the gradient (Slope " m ") of line q ;



Now, since we already know the gradient let's find of the equation of line by using its Slope and one of the points using point slope form of line :


Now, plug in the value of gradient ~

here we can clearly observe that, the Area under the curve can easily be represented as :

Since, all the values of y that lies in the shaded region is smaller than the actual value of y for the corresponding values of x in the equation of line q
The factory made 3 batches of cookies.
(See my work below)
Answer:
x = -3 +/- square root(22)
Step-by-step explanation:
x = -b +/- square root(b^2 - 4ac) / 2a
ax^2 + bx + c = 0
these are both the quadratic formula but one is solved for the x and another for 0
a= 1
b= 6
c = -13
x= -6 +/- square root( 6^2 - 4(1)(13)) / 2(1)
x = -6 +/- sqrt( 36 + 52) / 2
x= -6 +/- sqrt (88) / 2
sqrt of 88 = 2 x sqrt (22)
divide 2 on each
x= -3 +/- sqrt (22)