Answer:
12 3/8
Step-by-step explanation:
16 -3 5/8 =12 3/8
to check:
12+3=15,
3/8+5/8= 8/8 which =1
so, 15+1=16
C^2+7 would equal 88. the first equation states that C=9, so you can just plug in 9 for C
So 1/50 brought the pass
x out 1500
1/50 = x/1500
cross multiply
1500 = 50x
divide both sides by 50
30 = x
You can expect for 30 out of the 1,500 to receive free admissions.
Hope this helps
As you progress in math, it will become increasingly important that you know how to express exponentiation properly.
y = 2x3 – x2 – 4x + 5 should be written <span>y = 2^x3 – x^2 – 4^x + 5. The
" ^ " symbol denotes exponentiation.
I see you're apparently in middle school. Is that so? If so, are you taking calculus already? If so, nice!
Case 1: You do not yet know calculus and have not differentiated or found critical values. Sketch the function </span>y = 2x^3 – x^2 – 4^x + 5, including the y-intercept at (0,5). Can you identify the intervals on which the graph appears to be increasing and those on which it appears to be decreasing?
Case 2: You do know differentiation, critical values and the first derivative test. Differentiate y = 2x^3 – x^2 – 4^x + 5 and set the derivative = to 0:
dy/dx = 6x^2 - 2x - 4 = 0. Reduce this by dividing all terms by 2:
dy/dx = 3x^2 - x - 2 = 0 I used synthetic div. to determine that one root is x = 2/3. Try it yourself. This leaves the coefficients of the other factor, (3x+3); this other factor is x = 3/(-3) = -1. Again, you should check this.
Now we have 2 roots: -1 and 2/3
Draw a number line. Locate the origin (0,0). Plot the points (-1, 0) and (2/3, 0). This subdivides the number line into 3 subintervals:
(-infinity, -1), (-1, 2/3) and (2/3, infinity).
Choose a test number from each interval and subst. it for x in the derivative formula above. If the derivative comes out +, the function is increasing on that interval; if -, the function is decreasing.
Ask all the questions you want, if this explanation is not sufficiently clear.
Answer:
8
Step-by-step explanation:
(Refer to picture)
Plug in the value of A and the value of B into the expression
