Answer: 56 1/4 % = 52.25 % = 52.25 /100 = 0.5225
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Answer:
$6,400
Step-by-step explanation:
9100-1100
8000
8000×1/5
$1600
8000-1600
$6,400
Answer:
y = x + 1
Step-by-step explanation:
The gradient of a line can be defined by the equation:
m (gradient) = (y1 – y2 ) ÷ (x1 – x2) ----> "1" and "2" should be in subscript
For (-7,-6) we use x2 and y2 (because this point can be anywhere along a line):
x2 = -7, y2 = -6
Plug these values into the formula above:
m = (y-(-6)) ÷ (x-(-7))
m = (y+6) ÷ (x+7)
At this stage, the equation can't be solved as there are two unknowns. Therefore, the gradient must be found another way. Two lines are parallel if they have the same gradient - in their y=mx+c equations, m will be equal.
x - y=7 is the line alluded to in the question. Rearranging this equation into the line equation format gives:
y = x-7 ---> The gradient (coefficient of x) is 1.
Therefore, the gradient of the other parallel line must also be 1.
This can be substituted into the previous equation to give:
1 = (y+6)÷(x+7)
x+7 = y+6
x+1 = y
Therefore, the answer is y=x+1
Answer:
y= -4
Step-by-step explanation:
x+y=42. We know that x plus y equals 42.
x=2y . We also know that x equals twice y.
2y+y=42. Since we know that x=2y, we're going to replace x with 2y from the original equation, simplified would be 3y=42.
y=14. 3y=42 simplified is y=14, this is because we divided both sides by 3 (since 3 and 42 are both divisible by 3) (btw divisible by 3 simply means that it can be divided by 3)
x+14=42. Now that we know what y equals, let's replace y with 14 from the original equation.
x=28. Now that we know that x=28 and y=14, we know that y is always x/2.
When x=-8, we divide it by 2 and get, -4.
Not absolutely positive as this was all mental math, but hope this helps! :)
Answer:
Step-by-step explanation:
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If One ball is drawn, its color is recorded, and it is replaced in the urn. Then another ball is drawn and its color is recorded.
The 25 Possible outcomes of this experiment are listed below:
The tree diagram of this event is also attached.
