Answer:
-16/13
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
m=(-9-7)/(17-4)
m=-16/13
9514 1404 393
Answer:
$2038.85
Step-by-step explanation:
The value of the loan at that point is given by ...
A = P(1 +rt) . . . . . Principal P, rate r, time t (years)
A = $1850(1 + 0.1225·(10/12)) = $2038.85
Ricardo will have paid back $2038.85 at the end of the loan period.
_____
<em>Additional comment</em>
We assume that the loan accrues simple interest and that the amount due is the sum of principal and interest at the end of the loan period.
The question is not specific as to whether interest compounds, or whether intermediate (monthly) payments are made. There are many possible ways the loan could be repaid, generally involving different amounts for the different terms.
Answer:
<h2>
y = ¹/₂x + 6</h2>
Step-by-step explanation:
The product of slopes of perpendicular lines is -1
y = - 2x + 6 ⇒ m₁ = - 2
m₁m₂ = -1
-2m₂ = -1
m₂ = ⁻¹/₋₂ = ¹/₂
The Point-Slope form of equation of a line with slope of "m" and passing through point (x₁, y₁) is: y - y₁ = m(x - x₁)
m = ¹/₂
(-2, 5) ⇒ x₁ = -2, y₁ = 5
So the equation in point-slope form:
y - 5 = ¹/₂(x - (-2))
Therefore:
y - 5 = ¹/₂x + 1
y = ¹/₂x + 1 + 5
<h3>
y = ¹/₂x + 6</h3>
Answer:
= -5/4
Step-by-step explanation:
The slope through two points can be found by
m = (y2-y1)/(x2-x1)
= (-4-1)/(3--1)
= (-4-1)/(3+1)
= -5/4
On September 25.2014 Colson Corp. sold 200.000 widgetrons for 5$ per unit.
200.000*5$=1.000.000$. Same day company was paid for 40% which is 400.000$.
The rest of the money he get paid in two equal parts on November 15.2014 and January 20.2015. So Calson Corp. was paid 300.000$ on November 15.2014 and 300.000$ on January 20.2015.
