The problem asks for a location that is equidistant from towns A and B and lies on the given road. Calculating the midpoint of A and B, we get (5, 1.5). However, this point does not lie on the road denoted by -x + 7y = -4. So, we cannot determine the exact location of the school with the given conditions.
In this problem, the location of the school should be the midpoint of the line between towns A and B as it is equidistant from both towns. First, let's calculate the midpoint (M) coordinates. The formulas for finding the x and y coordinates of the midpoint are (x1 + x2) / 2 and (y1 + y2) / 2 respectively. Using these formulas, we get the coordinates of M as (2+8)/2, (-2+5)/2 = (5, 1.5). However, we should ensure that this point lies on the given road, which is denoted by the equation -x + 7y = -4. Substituting the coordinates of M in the equation, we get -5 + 7*1.5 = -5 + 10.5 = 5.5 which is not equal to -4. So, (5, 1.5) is not a valid location for the school. Unfortunately, with the given conditions, we cannot determine the exact location of the school. Additional information or revision of the conditions might be necessary to solve this problem.
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-0.08x+0.10y=0.10
Step by step explanation please.
Answer:
I think it's the second formula.
Step-by-step explanation:
Let x be the original price.
The discount rate is 20%
Sale Price is $68.90
Discount = 20% of x = 0.2x
Using the formula, Sale Price = Original Price - Discount,
Original Price - Discount is x-0.2x = 0.8x
$68.90 = 0.8x
From here on, just do the equation :) Hope this helps!
B. y = –1x – 2
C. y = –1x – 1
D. y = –1x + 8
Answer:
Option B is correct.
Step-by-step explanation:
The distributive property says that:
Given the equation of line:
Apply the distributive property we have;
Add 3 to both sides we have;
Simplify:
Therefore, an equation describes the same line as y – 3 = –1(x + 5) is,
y – 3 = –1(x + 5)
y - 3 = -x - 5
y = -x - 5 + 3
y = -x - 2
answer is B. y = –1x – 2
X4-y4