math & puzzles Online Quiz - 173

Eagle, cloud, and plane are all things commonly found high in the sky or capable of flight. A squirrel is a terrestrial animal that, while it can climb trees, does not belong in the category of aerial objects or flyers.

The limit expression represents the definition of the derivative $f'(4)$. The problem provides values for $f'(0)$, $f'(2)$, and $f'(5)$, but does not provide the value for $f'(4)$. Since $f'(4)$ cannot be determined from the given information, 'None of the above' is correct.

The words 'now', 'cow', and 'how' all rhyme with each other (ending in the /aʊ/ sound). 'Low' is the odd one out because it has a different vowel sound (ending in the /oʊ/ sound), despite the similar spelling.

This is a pattern recognition question. November is the odd one out because August, December, and January are all consecutive months in the winter season (August-December in Northern Hemisphere academic calendars, or December-January as year transition). November stands alone in this grouping pattern.

The limit expression is the definition of the derivative of the function $H(x) = f(x) + g(x)$ at $x=1$. Therefore, the limit equals $f'(1) + g'(1)$. Substituting $x=1$ into $f'(x)=3x$ and $g'(x)=2x^2$ gives $3(1) + 2(1)^2 = 5$.

June has only 30 days, so '31st June' is a non-existent date. February 2000 was a leap year (29 days), April 1998 has 30 days (so 29th is valid), and August always has 31 days. The impossibility of the date makes it the odd one out.

Jacket, shirt, and trousers are specific types of finished garments. Cloth is the raw material or a general term for fabric used to make these garments, making it the odd one out in this category of specific apparel items.

A number is divisible by 4 if its last two digits ($ab$) form a number divisible by 4. Given $a + b = 11$ and $a < b$, possible pairs $(a,b)$ are $(2,9), (3,8), (4,7), (5,6)$. Among these, only $56$ is divisible by 4. Thus $a=5, b=6$ is the correct solution.

For a quadratic equation $ax^2 + bx + c = 0$, the sum of the reciprocals of the roots is $1/x_1 + 1/x_2 = (x_1 + x_2)/(x_1x_2) = (-b/a) / (c/a) = -b/c$. Here, $b = -3/5$ and $c = 11/3$. Thus, $-(-3/5) / (11/3) = (3/5) * (3/11) = 9/55$.

The expression $n! / (n - 2)!$ simplifies to $n(n - 1)$. We solve $n^2 - n - 342 = 0$. Factoring gives $(n - 19)(n + 18) = 0$. Since $n$ must be a positive integer, $n = 19$.

If $a, b$ are even, $a+1$ and $b+1$ are both odd. The ratio of two odd numbers can be an integer (e.g., $9/3 = 3$), and that integer will always be odd. The other options result in even numbers: (odd+odd=even), (odd*odd-1=even).

Solve the first equation: $2x + 4 = 4x - 8 \Rightarrow 12 = 2x \Rightarrow x = 6$. Substitute $x = 6$ into the second equation: $-6 + k = 2(6) - 1 \Rightarrow -6 + k = 11 \Rightarrow k = 17$.

Since a, b, c, and d are different numbers and abc = 0, one of them must be 0. Because a + b = d, if a or b were 0, then d would equal the other number, violating the 'different values' constraint. Therefore, c must be 0.

Using the Principle of Inclusion-Exclusion: Total students = $G + F + S - (GF + FS + GS) + GFS$. Total = $25 + 15 + 13 - (3 + 4 + 2) + 0 = 53 - 9 = 44$. Students not studying any = $80 - 44 = 36$.

$202^2 = (200 + 2)^2 = 40000 + 800 + 4 = 40804$. Rounding 40,804 to the nearest hundred gives 40,800, as the tens digit (0) is less than 5.

To solve this question, the user needs to know the concept of the work rate formula, which is given by:

work = rate x time

where work refers to the amount of work done, rate refers to the rate at which work is done, and time refers to the amount of time taken to complete the work.

Using this formula, we can derive the formula for this question as follows:

work = rate x time

x = wrn (1) # The amount of work done by w workers in n days to produce x toys
y = crt (2) # The amount of work done by c workers in t days to produce y toys

Since the workers are working at the same rate, we can equate the rates in equations (1) and (2):

wr = cr

r = w*c/r

Substituting r in equation (1), we have:

x = w * (w*c/r) * n

x = w^2 * c * n / r

Substituting r in equation (2), we have:

y = c * (w*c/r) * t

y = w*c^2 * t / r

Equating the value of r in both equations, we get:

wc/r = wc/r

Thus, we can equate the two expressions for r and solve for t:

wc/r = wc/r

wc/(wc) = n/t

t = nwc/(w*c) = n

Therefore, the formula for the number of days it takes c workers to produce y toys is:

y = w^2 * c * n / x

Simplifying this expression, we get:

ywn / x*c

Therefore, the answer is:

The Answer is: D

A remainder $< 10$ when divided by 100 means the tens digit 'a' must be 0. The number is $2130b$. The sum of digits is $2+1+3+0+b = 6+b$. If the sum is 13, then $6+b=13$, so $b=7$.

The equation $1/a + 1/|a| = 0$ implies $1/|a| = -1/a$. This is true for all negative values of $a$. If $a < 0$, then $|a| = -a$, so $1/(-a) = -1/a$, which is an identity. Thus, there are infinite solutions (all $a < 0$).

24 minutes and 40 seconds is $24 + 40/60 = 24.666...$ minutes. Distance = Speed $\times$ Time. $15 \text{ feet/min} \times 24.666... \text{ min} = 370 \text{ feet}$. Calculation: $15 \times 24 = 360$ and $15 \times (2/3) = 10$. $360 + 10 = 370$.

The function $f(x-3)$ is a horizontal shift of $f(x)$ to the right by 3 units. If the original domain is $[a, b]$, the new domain is found by solving $a \le x - 3 \le b$, which simplifies to $a + 3 \le x \le b + 3$.