CubeyThePenguin

numerator = f(g(f(2)))

= f(g(2 * 2 + 3))

= f(g(7))

= f(3 * 7 + 2)

= f(23) = 2 * 23 + 3 = 49

denominator = g(f(g(2)))

= g(f(3 * 2 + 2))

= g(f(8))

= g(2 * 8 + 3)

= g(19) = 3 * 19 + 2 = 59

The answer is 49/59.

Let the width of each flower bed be W and the length be L. (Here, I'm assuming

length:

5 + 4W = 23

4W = 18

W = 18/4 = 4.5 m.

width:

2 + L = 17

L = 15 m.

3/4 * 400 = 300 seniors major in four subjects, leaving 100 seniors for the other two subjects. We can allocate 100 - 40 = 60 seniors to any of the last two subjects, since a minimum of 20 seniors major in each subject.

That leaves 20 + 60 = 80 seniors in one subject and 20 seniors in the other.

\(g(x) = \frac{1}{x^3} = -10\)

\(1 = -10x^3 \qquad \rightarrow \qquad -\frac{1}{10} = x^3 \qquad \rightarrow \qquad x = \boxed{-\frac{1}{\sqrt[3]{10}}}\)

\(g(x) > f(x) \qquad \rightarrow \qquad \frac{1}{x^3} > x^2\)

\(1 > x^5 \qquad \rightarrow \qquad \boxed{x \in (-\infty, 1)}\)

g(x) = 1/(x^3)

asymptote at y = 0 and x = 0

f(x) = x^2

axis of symmetry at x = 0

Let the two numbers be x and y, with x being the larger number.

x - y = 18

3x + 2y = 74

Multiply the first equation by 3 so we can eliminate x.

3x - 3y = 54

-3x - 2y = -74

-------------------------

-5y = -20

y = 4 --> x = 22

The numbers are 22 and 4.

There are 4 nonworking flares and 8 working flares.

a) \(\frac{4 \choose 4}{12 \choose 4} = \boxed{\frac{1}{495}}\)

b) \(\frac{4 \choose 4}{12 \choose 4} + \frac{\binom{4}{3} \cdot \binom{8}{1}}{12 \choose 4} + \frac{\binom{4}{2} \cdot \binom{8}{2}}{12 \choose 4}= \boxed{\frac{67}{165}}\)

c) \(\frac{\binom{4}{4}}{\binom{12}{4}}(0) + \frac{\binom{4}{3} \cdot \binom{8}{1}}{\binom{12}{4}}(1) + \frac{\binom{4}{2} \cdot \binom{8}{2}}{\binom{12}{4}}(2) + \frac{\binom{4}{1} \cdot \binom{8}{3}}{\binom{12}{4}}(3) + \frac{\binom{8}{4}}{\binom{12}{4}}(4) = \frac{1320}{495} = \boxed{\frac{8}{3}} \)

Let L be the number of large boxes and S be the number of small boxes. We can write a system of equations to solve this problem:

L + S = 125

55L + 35S = 4950

Multiply the first equation by 55 so we can eliminate L:

55L + 55S = 6875

55L + 35S = 4950

---------------------------------------- (subtraction)

20S = 1925

S = 96.25

L = 125 - 96.25 = 28.75

I solved the problem exactly how it was supposed to, but the answers aren't integers, which doesn't make sense.

This question has been answered before: https://web2.0calc.com/questions/help-plz_55355

Otherwise, let's solve it here.

y = 1: x = 1, 2

y = 2: no solutions for x (you are solving 4x + 16 < 20, but x has to be less than 1)

So, there are only 2 pairs that solve this inequality.

On the calculator, you can use the parentheses to group expressions together in the numerator or the denominator.

For example, 2/3*6/9 gets (2/3 * 6)/9, but (2/3) * (6/9) would probably be the correct grouping to solve the problem.

When typing out answers, the LaTeX command for fractions is \frac{a}{b}, where a is the numerator and b is the denominator.

a) find the probability that 0 or 1 people have believed that they have seen a UFO, then subtract from 1.

\(1 - (0.1)^{0}(0.9)^{10} - (0.1)^{1}(0.9)^{9} \approx 0.6125 = \boxed{61.25\%}\)

b) \((0.1)^2(0.9)^8 + (0.1)^3(0.9)^7 \approx 0.00478 = \boxed{0.478\%}\)

c) \((0.1)^1(0.9)^9 \approx 0.0387 = \boxed{3.87\%}\)