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VMariaS [17]
3 years ago
15

The equation of the line EF is y=2x+1. Write an equation of a line parallel to line EF in slope intercept form that contains poi

nt 0,2
Mathematics
2 answers:
saul85 [17]3 years ago
5 0
Y = 2x + 2

The slope must stay the same since they are parallel and the new point would be the y intercept. 
Ymorist [56]3 years ago
5 0
First, know that - 
· slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept
· parallel lines have the same slope

Given the above, the slope of line EF is 2. This means the equation of the line parallel to line EF will also have a slope of 2 (parallel lines share a slope). If we plug in our slope, the slope-intercept equation becomes y = 2x + b.

Now, we have to find b, the y-intercept. Since the line must pass through the point (0, 2), we can plug it in for the (x, y) values in our equation and solve algebraically for b.

    y = 2x + b
    2 = 2(0) + b
    2 = 0 + b
    2 = b

The y-intercept is 2. We now have the complete equation in slope-intercept form parallel to y = 2x + 1 which passes through the point (0, 2)

Answer:
y = 2x + 2
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Answer:

5 trees should be planted to maximize the yield per acre,

The maximum yield would be 1250

Step-by-step explanation:

Given,

The original number of trees per acre = 20,

Average pounds of nuts by a tree = 60,

Let x be the times of increment in number of trees,

So, the new number of trees planted per acre = 20 + x

∵ for each additional tree planted per acre, the average yield per tree drops 2 pounds,

So, the new number of pounds of nut = (60 - 2x)

Thus, the total yield per acre,

Y(x) = (20+x)(60-2x)

Differentiating with respect to t ( time ),

Y'(x) = (20+x)(-2) + 60 - 2x = -40 - 2x + 60 - 2x = 20 - 4x

Again differentiating with respect to t,

Y''(x) = -4

For maxima or minima,

Y'(x) = 0

⇒ 20 - 4x = 0

⇒ 20 = 4x

⇒ x = 5,

For x = 5, Y''(x) = negative,

Hence, Y(x) is maximum for x = 5,

And, maximum value of Y(x) = (20+5)(60 - 10) = 25(50) = 1250,

i.e. 5 trees should be planted to maximize the yield per acre,

and the maximum yield would be 1250 pounds

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beks73 [17]
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Answer:

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