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PolarNik [594]
3 years ago
13

What is a necessary step for constructing perpendicular lines through a point off the line?

Mathematics
2 answers:
mixas84 [53]3 years ago
7 0

Draw arcs on either side of a given point on the line

To construct a line perpendicular to another line through a point on the line:

Measure a length from that point to one edge of the line

Draw two intersecting arcs on either sides of the line

Draw a straight line from the intersecting point of the arcs to the point on the existing line.

The new line drawn is a perpendicular line.

The only step necessary out of the given choices is "Draw arcs on either side of a given point on the line"

Option C is the correct choice

Learn more here: brainly.com/question/929137

Andrej [43]3 years ago
5 0

Answer:

was it c?

Step-by-step explanation:

have a great day

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60

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It is just 60 I did my work lol

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5

Step-by-step explanation:

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3 years ago
Dairy farmers are aware there is often a linear relationship between the age, in years, of a dairy cow and the amount of milk pr
Jobisdone [24]

Answer:

B. A cow of 5 years is predicted to produce 5.5 more gallons per week.

Step-by-step explanation:

Let M(a) = 40.8-1.1\cdot a, where a is the age of the dairy cow, measured in years, and M(a) is the predicted milk production, measured in gallons per week.

Besides, we consider a_{1} and a_{2}, such that a_{1}\ne a_{2}, we define the difference between predicted milk productions (\Delta M) below:

\Delta M = -1.1\cdot (a_{2}-a_{1}) (1)

If we know that a_{1} = 5\,yr and a_{2} = 10\,yr, then the difference between predicted milk productions is:

\Delta M = -1.1\cdot (10-5)

\Delta M = -5.5\,\frac{gal}{week}

That is, a cow of 5 years is predicted to produce 5.5 more gallons per week than a cow of 10 years. Hence, the right answer is B.

6 0
2 years ago
What is the answer to this calculus problem???<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%7D%7Bdx%7D%20%28e%20%7B%7D%5E
Len [333]

Answer:

see explanation

Step-by-step explanation:

Differentiate using the product rule

Given y = f(x)g(x), then

\frac{dy}{dx} = f(x). g'(x) + g(x). f'(x)

here f(x) = e^{x} ⇒ f'(x) = e^{x}

g(x) = cosx ⇒ g'(x) = - sinx

Hence

\frac{dy}{dx} = e^{x}(- sinx) + cosx e^{x}

                                   = e^{x}cosx - e^{x}sinx

3 0
2 years ago
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