First of all we have to find the slope
m = y₂-y₁ / x₂-x₁
m = -8 + 16 / -10 + 8
m = 8/-2
m = -4
y-y₁ = m (x-x₁)
y + 8 = -4 (x + 10)
y + 8 = -4x - 40
y = -4x -40 - 8
y = -4x -48
Answer:
10. 7n - 1 < -8
Isolate the variable, n. Do the opposite of PEMDAS. Treat the < as equal sign, what you do to one side, you do to the other. First, add 1 to both sides:
7n - 1 (+1) < - 8 (+1)
7n < - 8 + 1
7n < - 7
Isolate the variable, n. Divide 7 from both sides:
(7n)/7 < (-7)/7
n < -7/7
n < -1
n < -1 is your answer.
11. 3 > -7v + 4v
Combine like terms, then isolate the variable, v. First, add -7v and 4v together.
3 > (-7v + 4v)
3 > (4v - 7v)
3 > (-3v)
Isolate the variable, v. Divide -3 from both sides. Note that since you are dividing a negative number, you must flip the sign:
(3)/-3 > (-3v)/-3
3/-3 > v
-1 < v
v > -1 is your answer.
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If 2 equations have the same y-intercept, they are overlapping, which means they have infinite solutions. So there is no way that 2 equations with the same y-intercept will have no solution. Thus your answer is: C)Never.
The first year, the amount is 40,000
the second year is 40000 + 4.2% of 40000, or 0.042 * 4000, so 40000+(0.042*4000)
common factoring that we get 40000(1 + 0.042), or just 40000(1.042)
in short, the starting amount is 40000, and to get the next term's value you'd use the "common ratio" of 1.042, namely the multiplier of 1.042.
for the third year it'll be 40000(1.042) + (0.042 *
40000(1.042) ), again, common factoring that
40000(1.042)(1 + 0.042) or 40000(1.042)(1.042) or 40000(1.042)²
therefore,
First, you need to get delta which is b squared minus 4ac. If delta is higher than cero, the polynomial has two solutions, if it is less, the polynomial has no real solutions and if it is the same, it has one solution.
Although that is only for second grade ecuations