y=x^2 + x - 2
x + y = 1
Replace y in the second equation:
x + x^2 + x -2 = 1
Simplify:
x^2 + 2x -2 = 1
Subtract 1 from both sides:
x^2 + 2x -3 = 0
Factor:
(x-1) (x+3) = 0
Solve for both x's:
x = 1 and x = -3
Now replace x in the second equation and solve for y using both x values:
1 + y = 1, y = 0
-3 + y = 1, y = 4
Now you have (1,0) and (-3,4) as solutions for (x,y)
XY = x times y:
1 x 0 = 0
-3 x 4 = -12
The answer would be -12
Answer:
Y(n) = 7n + 23
Step-by-step explanation:
Given:
f(0) = 30
f(n+1) = f(n) + 7
For n=0 : f(1) = f(0) + 7
For n=1 : f(2) = f(1) + 7
For n=2 : f(3) = f(2) + 7 and so on.
Hence the sequence is an arithmetic progression with common difference 7 and first term 30.
We have to find a general equation representing the terms of the sequence.
General term of an arithmetic progression is:
T(n) = a + (n-1)d
Here a = 30 and d = 7
Y(n) = 30 + 7(n-1) = 7n + 23
First, change each mixed fraction into improper
2 3/8 = 19/8
1 3/4 = 7/4
The problem will look like: (19/8)/(7/4)
To solve, flip the second fraction, and change the division sign into a multiplication sign.
(19/8)/(7/4) = (19/8) x (4/7)
Multiply across
(19/8) x (4/7) = 76/56
Simplify.
76/56 simplified = 19/14
Change to mixed fraction
19/14 = 14/14 + 5/14 = 1 5/14
1 5/14 is your answer
hope this helps
Answer:
2. 5 miles
Step-by-step explanation:
Given that:
Rental cost = $30
Cost per mile driven = 18 cent
Daily budget = $75
Greatest distance you can drive each day while staying within budget :
Rental cost + (cost per mile driven * distance) ≤ daily budget
$30 + 18d ≤ $75
18d ≤ 75 - 30
18d ≤ 45
d ≤ 45 / 18
d ≤ 2.5
Greatest distance you can drive each day while staying within budget is 2.5 miles.
