To determine which of the given options divides 726179, we can use the division method. We will divide 726179 by each option and check if the remainder is zero. If the remainder is zero, then the number is divisible by that option.
A) 23 — As shown in the step-by-step calculation, when 726179 is divided by 23, the remainder is 0. Therefore, 726179 is divisible by 23.
To determine the type of quadrilateral, we need to check its properties based on the given angles. Specifically, we will check if it satisfies the conditions for a cyclic quadrilateral, trapezium, rhombus, or parallelogram.
C) Cyclic quadrilateral. A quadrilateral is cyclic if and only if the sum of each pair of its opposite angles is \(180^\circ\). As calculated, \(\angle A + \angle C = 180^\circ\) and \(\angle B + \angle D = 180^\circ\).
To find the surface area of a sphere, we use the standard formula for the surface area of a sphere, which is \(4 \pi r^2\), where \(r\) is the radius of the sphere. We are given the radius, so we just need to substitute it into the formula and calculate the result.
Thus, the surface area of the sphere is \(2304 \pi \text{ cm}^2\).
Correct Option: C) 2304 \pi cm²
This problem involves setting up and solving a system of two linear equations with two variables. We will represent the mother's age and the daughter's age with variables, translate the given conditions into equations, solve the system of equations to find their individual ages, and then calculate the difference between their ages.
Let \(M\) be the mother's age in years.
Let \(D\) be the daughter's age in years.
"3 times the mother's age is 29 years more than 5 times her daughter's age"
This can be written as:
\[ 3M = 5D + 29 \quad \text{(Equation 1)} \]"2 times the daughter's age is 2 years less than the mother's age"
This can be written as:
\[ 2D = M - 2 \]Rearranging this equation to express \(M\) in terms of \(D\):
\[ M = 2D + 2 \quad \text{(Equation 2)} \]Substitute the expression for \(M\) from Equation 2 into Equation 1:
\[ 3(2D + 2) = 5D + 29 \]Distribute the 3:
\[ 6D + 6 = 5D + 29 \]Subtract \(5D\) from both sides:
\[ 6D - 5D + 6 = 29 \] \[ D + 6 = 29 \]Subtract 6 from both sides to find \(D\):
\[ D = 29 - 6 \] \[ D = 23 \]So, the daughter's age is 23 years.
Substitute the value of \(D\) back into Equation 2:
\[ M = 2(23) + 2 \] \[ M = 46 + 2 \] \[ M = 48 \]So, the mother's age is 48 years.
Difference \( = M - D \)
\[ \text{Difference} = 48 - 23 \] \[ \text{Difference} = 25 \]A) 25
To find the Least Common Multiple (LCM) of given numbers expressed in prime factorization form, we first need to express all numbers as a product of their prime factors. Then, for each prime factor, we take the highest power that appears in any of the given numbers. The product of these highest powers will be the LCM.
The given numbers are:
Let's convert all composite bases to their prime factors:
Now, substitute these into the given numbers:
The prime factors involved are 2, 3, and 13.
Multiply the highest powers of all unique prime factors:
\[ \text{LCM} = 2^2 \times 3^{10} \times 13^2 \]Correct Option: B) \(2^2 \times 3^{10} \times 13^2\)
Simplify \(\frac{(x^2-9)(9x^2-1)}{\left(x-\frac{1}{3}\right)\left(1-\frac{x}{3}\right)}\times\frac{9}{\left(x+\frac{1}{3}\right)\left(1+\frac{x}{3}\right)}\).
The problem requires simplifying a complex algebraic expression involving fractions and products. The key is to factorize the terms using algebraic identities such as \(a^2 - b^2 = (a-b)(a+b)\) and to simplify the denominators by taking common factors.
Let the given expression be \(E\).
\[ E = \frac{(x^2-9)(9x^2-1)}{\left(x-\frac{1}{3}\right)\left(1-\frac{x}{3}\right)}\times\frac{9}{\left(x+\frac{1}{3}\right)\left(1+\frac{x}{3}\right)} \]Factorize the numerators using the identity \(a^2 - b^2 = (a-b)(a+b)\):
So the numerator becomes \((x-3)(x+3)(3x-1)(3x+1)\).
Simplify the denominators:
Substitute these simplified terms back into the expression:
\[ E = \frac{(x-3)(x+3)(3x-1)(3x+1)}{\left(\frac{3x-1}{3}\right)\left(-\frac{x-3}{3}\right)}\times\frac{9}{\left(\frac{3x+1}{3}\right)\left(\frac{x+3}{3}\right)} \]Combine the denominators in the first fraction:
\[ \left(\frac{3x-1}{3}\right)\left(-\frac{x-3}{3}\right) = -\frac{(3x-1)(x-3)}{9} \]Combine the denominators in the second fraction:
\[ \left(\frac{3x+1}{3}\right)\left(\frac{x+3}{3}\right) = \frac{(3x+1)(x+3)}{9} \]Substitute these back into the expression:
\[ E = \frac{(x-3)(x+3)(3x-1)(3x+1)}{-\frac{(3x-1)(x-3)}{9}}\times\frac{9}{\frac{(3x+1)(x+3)}{9}} \]Rewrite the division by fractions as multiplication by reciprocals:
\[ E = (x-3)(x+3)(3x-1)(3x+1) \times \left(-\frac{9}{(3x-1)(x-3)}\right) \times 9 \times \left(\frac{9}{(3x+1)(x+3)}\right) \]Cancel out common terms:
After cancellation, the expression simplifies to:
\[ E = 1 \times \left(-9\right) \times 9 \times 9 \]Calculate the final product:
\[ E = -9 \times 81 = -729 \]A) -729 is the correct simplification of the given expression.
This problem involves the relationship between the H.C.F. (Highest Common Factor), L.C.M. (Least Common Multiple), and two numbers. The fundamental property states that the product of two numbers is equal to the product of their H.C.F. and L.C.M.
Correct Option: B) 200
To find the length of the bridge, we need to use the relationship between speed, distance, and time. The car's speed is given in km/hr, and the time taken to cross the bridge is given in minutes. We must ensure that all units are consistent before performing calculations. It's often easiest to convert everything to SI units (meters and seconds) or to kilometers and hours.
Since the speed is in km/hr, we should convert the time from minutes to hours.
\[ T = 4.6 \text{ minutes} = \frac{4.6}{60} \text{ hours} \]The distance covered by the car is equal to the length of the bridge. The formula relating speed, distance, and time is:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]Substitute the values into the formula:
\[ \text{Length of bridge} = 48 \text{ km/hr} \times \frac{4.6}{60} \text{ hours} \]\[ \text{Length of bridge} = \frac{48 \times 4.6}{60} \text{ km} \]\[ \text{Length of bridge} = \frac{220.8}{60} \text{ km} \]\[ \text{Length of bridge} = 3.68 \text{ km} \]Thus, the length of the bridge is 3.68 km.
D) 3.68 km is the correct answer, as calculated by converting the time to hours and multiplying by the speed.
We are given two conditions: the product of two positive numbers is 2000, and the first number is five times the second number. We need to find the sum of these two numbers. We will set up a system of two equations with two variables and solve for the numbers, then find their sum.
Let the two positive numbers be \(x\) and \(y\).
From the first condition, "The product of two positive numbers is 2000", we can write the equation:
\[ xy = 2000 \quad (1) \]From the second condition, "If the first number is five times the second number", we can write the equation:
\[ x = 5y \quad (2) \]Now, substitute the expression for \(x\) from equation (2) into equation (1):
\[ (5y)y = 2000 \]\[ 5y^2 = 2000 \]Solve for \(y\):
\[ y^2 = \frac{2000}{5} \]\[ y^2 = 400 \]\[ y = \sqrt{400} \]Since the numbers are positive, we take the positive square root:\[ y = 20 \]Now, substitute the value of \(y\) back into equation (2) to find \(x\):
\[ x = 5y \]\[ x = 5(20) \]\[ x = 100 \]The two numbers are 100 and 20. We need to find their sum:
\[ \text{Sum} = x + y \]\[ \text{Sum} = 100 + 20 \]\[ \text{Sum} = 120 \]A) 120 is the correct sum of the two numbers.
Pipe P can fill \( \frac{2}{7} \) part of a tank in 18 hours, and pipe Q can fill \( \frac{5}{9} \) part of the same tank in 25 hours. Both P and Q were kept open for 3 hours, then both were closed. Pipe R alone was then opened and it emptied the water in the tank in 8 hours. Pipes P, Q and R together can fill the empty tank in:
To solve this problem, we need to determine the individual filling rates of pipes P and Q, and the emptying rate of pipe R. Then, we can calculate the net filling rate when all three pipes P, Q, and R work together.
Calculate the filling rate of Pipe P:
Pipe P fills \( \frac{2}{7} \) of the tank in 18 hours.
So, Pipe P fills the entire tank in:
\[ 18 \text{ hours} \div \frac{2}{7} = 18 \times \frac{7}{2} = 9 \times 7 = 63 \text{ hours} \]Calculate the filling rate of Pipe Q:
Pipe Q fills \( \frac{5}{9} \) of the tank in 25 hours.
So, Pipe Q fills the entire tank in:
\[ 25 \text{ hours} \div \frac{5}{9} = 25 \times \frac{9}{5} = 5 \times 9 = 45 \text{ hours} \]Calculate the amount of tank filled by P and Q in 3 hours:
In 1 hour, Pipe P fills \( \frac{1}{63} \) of the tank.
In 1 hour, Pipe Q fills \( \frac{1}{45} \) of the tank.
In 1 hour, P and Q together fill:
\[ \frac{1}{63} + \frac{1}{45} = \frac{5}{315} + \frac{7}{315} = \frac{12}{315} = \frac{4}{105} \text{ of the tank} \]In 3 hours, P and Q together fill:
\[ 3 \times \frac{4}{105} = \frac{4}{35} \text{ of the tank} \]Calculate the emptying rate of Pipe R:
Pipe R empties the \( \frac{4}{35} \) part of the tank in 8 hours.
So, Pipe R alone can empty the entire tank in:
\[ 8 \text{ hours} \div \frac{4}{35} = 8 \times \frac{35}{4} = 2 \times 35 = 70 \text{ hours} \]Calculate the combined filling rate of P, Q, and R:
Pipe P fills \( \frac{1}{63} \) of the tank per hour.
Pipe Q fills \( \frac{1}{45} \) of the tank per hour.
Pipe R empties \( \frac{1}{70} \) of the tank per hour.
When all three pipes work together, the net filling rate per hour is:
\[ \frac{1}{63} + \frac{1}{45} - \frac{1}{70} \]To add/subtract these fractions, find the Least Common Multiple (LCM) of 63, 45, and 70.
Prime factorization: \( 63 = 3^2 \times 7 \), \( 45 = 3^2 \times 5 \), \( 70 = 2 \times 5 \times 7 \).
LCM \( = 2 \times 3^2 \times 5 \times 7 = 2 \times 9 \times 5 \times 7 = 630 \).
Net filling rate:
\[ \frac{10}{630} + \frac{14}{630} - \frac{9}{630} = \frac{10 + 14 - 9}{630} = \frac{24 - 9}{630} = \frac{15}{630} \]Simplify the fraction:
\[ \frac{15}{630} = \frac{3 \times 5}{126 \times 5} = \frac{3}{126} = \frac{1}{42} \]So, P, Q, and R together fill \( \frac{1}{42} \) of the tank per hour.
Calculate the time taken to fill the empty tank:
If they fill \( \frac{1}{42} \) of the tank in 1 hour, they will fill the entire tank in 42 hours.
Correct Option: B) 42 hours