1. Cross multiply
35x = 5(11)
35x = 55
Divide both sides by 35
x = 55/35
x = 11/7

2. (x - 2)/x = 3/8
Cross multiply
3x = 8(x - 2)
3x = 8x - 16
Subtract 8x from both sides
-5x = -16
divide both sides by -5
x = -16/-5
x = 16/5 OR 3 1/5

3. (a + 1)/(a - 1) = 5/6
cross multiply
6(a + 1) = 5(a - 1)
distribute
6a + 6 = 5a - 5
subtract 5a from both sides
a + 6 = -5
subtract 6 from both sides
a = -11

4. (1/3)x - 4 = (2/3)x + 6
multiply each term by 3 to clear the fractions
x - 12 = 2x + 18
subtract x from both sides
-12 = x + 18
subtract 18 from both sides
-30 = x

3 0

4 years ago

Two children own two-way radios that have a maximum range of 3 miles. One leaves a certain point at 1:00 P.M., walking due north

Grace [21]

Answer:

The answer is 21 minutes

Step-by-step explanation:

We use the equation Xf = Xo + vt

1) At 1:00 PM, child one leaves the starting point heading north at a constant velocity of 6 mi/hr or .1 [mi/min] (divide by 60 to convert from [mi/hr] to [mi/min])

2) He walks for 15 minutes before kid 2 starts walking. In 15 minutes he is able to cover 1.5 [mi]

$x_{1f1} =x_{o} +v_{1} t_{1} \\x_{1f1} =0+.1(15)\\x_{1f1} =1.5 [mi]$

3) Now, child 2 starts walking and we know that when the range reaches 3 miles, they won´t be able to communicate. So the sum of the final position of child 1 and child 2 must be 3[mi]