Let the distance of Judy from intersection is x and distance of Jackie from intersection is y.
We convert the given information to equations, step by step.
Point 1: When Jackie is 1 mile farther from the intersection than Judy.
This means y is 1 mile more than x.
So,
y = 1 + x
Point 2: The distance between them is 2 miles more than Judy’s distance from the intersection.
Distance between is x+ y.
So, x+y is 2 miles more than y.
x+y = y + 2
⇒
x = 2
From point 1 we have:
y = 1 + x = 1+ 2 = 3
So,
Distance of Judy from intersection is 2 miles and distance of Jackie from intersection is 3 miles.
Answer:
Pam: $181
Amanda: $362
Julie: $452
Step-by-step explanation:
(What does Mike have to do with this problem?)
Let a = Amanda's pay
Let p = Pam's pay
Let j = Julie's pay
"Amanda made twice what Pam earned"
a = 2p
"Julie made $90 more than Amanda"
j = a + 90
j = 2p + 90
Pam earned p
Total salary
a + p + j = 2p + p + 2p + 90
Total salary
$995
2p + p + 2p + 90 = 995
5p = 905
p = 181
a = 2p = 2(181) = 362
j = 2p + 90 = 362 + 90 = 452
Answer:
Pam: $181
Amanda: $362
Julie: $452
Given the table
C : 1, 2, 3, 4, 5
F(C) : 4, 6, 8, 10, 12
Usually, the independent variable is represented by the x-value while the dependent variable (or the function) is represented by the y-value.
Thus the points represented by the table are:
(1, 4), (2, 6), (3, 8), (4, 10) and (5, 15).
ST/VT = UT/WT6+14/14 = (x+4)+21/21X= 20*21/14 - 21x=420/14-25X=30-25x=5
The answer would be 34 12/100 or 34 3/50