A=number of seats in section A
B=number of seats in section B
C=number of seats in section C
We can suggest this system of equations:
A+B+C=55,000
A=B+C ⇒A-B-C=0
28A+16B+12C=1,158,000
We solve this system of equations by Gauss Method.
1 1 1 55,000
1 -1 -1 0
28 16 12 1,158,000
1 1 1 55,000
0 -2 -2 -55,000 (R₂-R₁)
0 12 16 382,000 (28R₁-R₂)
1 1 1 55,000
0 -2 -2 -55,000
0 0 4 52,000 (6R₂+R₃)
Therefore:
4C=52,000
C=52,000/4
C=13,000
-2B-2(13,000)=-55,000
-2B-26,000=-55,000
-2B=-55,000+26,000
-2B=-29,000
B=-29,000 / -2
B=14,500.
A + 14,500+13,000=55,000
A+27,500=55,000
A=55,000-27,500
A=27,500.
Answer: there are 27,500 seats in section A, 14,500 seats in section B and 13,000 seats in section C.
Solve out
x^2 + 64 = 0
x^2 = -64 ( we are going to use imaginary numbers to solve)
x =
x = 8i, -8i final answers
Please note that your x^3/4 is ambiguous. Did you mean (x^3) divided by 4
or did you mean x to the power (3/4)? I will assume you meant the first, not the second. Please use the "^" symbol to denote exponentiation.
If we have a function f(x) and its derivative f'(x), and a particular x value (c) at which to begin, then the linearization of the function f(x) is
f(x) approx. equal to [f '(c)]x + f(c)].
Here a = c = 81.
Thus, the linearization of the given function at a = c = 81 is
f(x) (approx. equal to) 3(81^2)/4 + [81^3]/4
Note that f '(c) is the slope of the line and is equal to (3/4)(81^2), and f(c) is the function value at x=c, or (81^3)/4.
What is the linearization of f(x) = (x^3)/4, if c = a = 81?
It will be f(x) (approx. equal to)
The more painstaking way would be to divide the spinner up into 100 pieces, and shade in 5 of those pieces. Since 5/100 = 0.05 = 5% this means that landing on any of the 5 pieces has a 5% chance of happening.
A better way to do this is to use 20 slices instead. This works because 5/100 reduces to 1/20 (divide top and bottom by 5). Instead of shading in 5 slices, Mark should shade in 1. If you use a calculator, you should see that 1/20 = 0.05 = 5% as well.
The second option is more practical but even then a spinner with 20 sections is still a lot. I think it might be better for Mark to pull random numbers out of a hat, or let the computer generate random numbers.
Hi there!
From the diagram, we can see that ΔAEF, and ΔACB are similar, isosceles triangles.
Therefore:
∠B and ∠F are congruent because they are corresponding angles, and ∠B and ∠E are congruent because of equal base angles in an isosceles triangle.
Therefore:
7x - 11 = 6x
Subtract 6x from both sides:
x - 11 = 0
Add 11 to both sides:
x = 11.
