1. It takes Dustin 2 hours to shovel the snow from his driveway and sidewalk. It takes his sister 3 hours to shovel the same area. How long will it take them to shovel the walk if they work together?
A: Working together, Dustin and his sister can shovel the walk in 1 1/5 hours, or 1 hour and 12 minutes?
2. A chemistry student needs to make a solution that is 70% water and 30% hydrochloric acid. The student's current mixture of 300 ml is 60% water and 40% hydrochloric acid. How much water must the student add to achieve his desired solution?
A: ?
Solve. Check your answer.
3. ((d + 2)/(d + 8)) = ((-6)/(d + 8))
A: ?
4. 4/n^2 = 7/n + 2
A: n = -4 or n = 1/2?
5. 2/t + 4/3t = 4/t + 2
A: t = 10?
6. ((x -6)/(x^2-6)) = ((-4)/(x-4))
A: ?
look at problem 5, it helps on the first two.
5.
2/t + 4/3t = 4/t + 2
multiply each side by 3t
6+4=12 + 6t
6t=-2
solve for t.
Now number one
time=work/rate
Now consider in 6 hours, D could do 3 walkwasy, and sister two walkways. so in six hours, they could shovel 5 walkeways, or combined rate= 5/6 walks/hr
time= 1walkwaytodo/(5/6 walway/hr)=6/5 hr
#3
no need for all those brackets ...
(d + 2)/(d + 8) = (-6)/(d + 8)
multiply both sides by (d+8)
d+2 = -6
d = -8
ahhh, trick question.
Notice in the original, d≠ -8
So there is no solution.
#4 correct
#5 handled by bobpursley
#6
(x -6)/(x^2-6) = (-4)/(x-4)
x^2 - 10x + 24 = -4x^2 + 24
5x^2 - 10x = 0
x^2 - 2x = 0
x(x-2) = 0
x = 0 or x = 2
What about #1 and #2?
Also, is #5 correct? bobpursley's response was somewhat unclear.
Reiny?
2. To find out how much water the student needs to add, we can set up an equation to represent the problem. Let's denote the amount of water the student needs to add as x.
The student's current mixture of 300 ml is 60% water, so it contains 0.6 * 300 = 180 ml of water.
The desired solution needs to be 70% water, so the total amount of solution should be 300 + x ml.
To achieve the desired solution, the amount of water in the final mixture should be 0.7 times the total amount of solution.
Therefore, we can set up the equation:
0.7 * (300 + x) = 180
To solve this equation for x, we can begin by dividing both sides by 0.7:
300 + x = 180 / 0.7
Next, we subtract 300 from both sides:
x = (180 / 0.7) - 300
Now we can use a calculator to find the value of x.
x ≈ 257.14 ml
So, the student needs to add approximately 257.14 ml of water to achieve the desired solution.
3. To solve the equation ((d + 2)/(d + 8)) = ((-6)/(d + 8)), we can start by multiplying both sides of the equation by (d + 8) to eliminate the denominators:
(d + 2) = -6
Next, we can subtract 2 from both sides to isolate the variable d:
d = -6 - 2
Simplifying further,
d = -8
So, the solution to the equation is d = -8.
6. To solve the equation ((x - 6)/(x^2 - 6)) = ((-4)/(x - 4)), we can start by cross-multiplying:
(x - 6)(x - 4) = (-4)(x^2 - 6)
Expanding both sides of the equation:
(x^2 - 4x - 6x + 24) = -4x^2 + 24
Combining like terms:
x^2 - 10x + 24 = -4x^2 + 24
Rearranging the equation:
x^2 + 4x^2 - 10x - 24 + 24 = 0
5x^2 - 10x = 0
Factoring out the common factor x:
x(5x - 10) = 0
Setting each factor equal to zero and solving for x:
x = 0 or 5x - 10 = 0
For the second equation, let's isolate x:
5x - 10 = 0
Adding 10 to both sides:
5x = 10
Dividing both sides by 5:
x = 2
So, the solutions to the equation are x = 0 and x = 2.