Answer:
Then the solution is (4, 6).
Step-by-step explanation:
Let's use the substitution method:
First multiply the second equation by 4, obtaining 4y = -2x + 32.
Now substitute (-2x + 32) for 4y in the first equation:
3x + (-2x + 32) = 36, or
3x - 2x + 32 = 36. or
x = 4.
If x = 4, then the second equation yields y = (-1/2)(4) + 8, or
y = -2 + 8, or y = 6
Then the solution is (4, 6).
Check, using the first equation:
Does 3(4) + 4(6) = 36? Does 12 + 24 = 36? YES
Answer:
t=0 or t=44
Step-by-step explanation:
There are a few methods to do this, the most obvious being the use of the quadratic formula.
However, a quick check with a calculator or even your mind will reveal that 704 is divisible by 16. Thus we can solve this through factoring.
-16t(t-44)=0
Using the zero factor principle:
-16t=0 and so t=0
t-44=0 and so t=44
And thus we get the solution.
Answer:
f(x) > 0 over the interval 
Step-by-step explanation:
If f(x) is a continuous function, and that all the critical points of behavior change are described by the given information, then we can say that the function crossed the x axis to reach a minimum value of -12 at the point x=-2.5, then as x increases it ascends to a maximum value of -3 for x = 0 (which is also its y-axis crossing) and therefore probably a local maximum.
Then the function was above the x axis (larger than zero) from
, until it crossed the x axis (becoming then negative) at the point x = -4. So the function was positive (larger than zero) in such interval.
There is no such type of unique assertion regarding the positive or negative value of the function when one extends the interval from
to -3, since between the values -4 and -3 the function adopts negative values.
Using a table of values, the outputs of f(x) for whole numbers are 0, 1, 4, 9, 16, 25, 36, and so on. For the same input values, g(x) has outputs of 1, 2, 4, 8, 16, 32, and 64. Continuing to double the output each time results in larger outputs than those of f(x). The exponential function, g(x), has a constant multiplicative rate of change and will increase at a faster rate than the quadratic function.
(ed. just click all of them)
I did not get any zeros since the graph doesn’t cross the x axis, meaning that there are no rational zeros
However, here is the method u can use to find the zeros lol
You can use the quadratic formula in order to get the zeros
This is the equation therefore use these values
ax^2+bx+c=0
A=1
B= -5
C=12
The quadratic formula is -b±√(b^2-4ac))/2a (I left a picture just in case)