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kupik [55]
3 years ago
11

A rectangular garden has the length of 12 feet. You need 36 feet of fencing to enclose the garden.what is the width of the garde

n?
Mathematics
2 answers:
NISA [10]3 years ago
8 0

Answer:

6

Step-by-step explanation:

Remember that there are 4 sides to a rectangle.

(2 x length) + (2 x width) = perimeter

(2 x 12) + (2 x width) = 36

24 + (2 x width) = 36

2 x width = 12

width = 6



Xelga [282]3 years ago
6 0

Answer:

Width = 6 Feet

Step-by-step explanation:

Length - 12 Feet

Whole Garden - 36 Feet


Use Division:

36 / 12

= 3

There are 2 sides to the width and length

3 * 2 = 6

Best Of Luck,

- I.A. -

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tresset_1 [31]

Plot the point (-7, -5). We are in quadrant 3.

We also know that tan θ = opp/adj.

Cot θ = adj/opp.

Let us use a^2 + b^2 = r^2 to show you how to find r, the radius aka hypotenuse.

Look: (-7)^2 + (-5)^2 = r^2

If you should ever need to find r, do this:

(-7)^2 + (-5)^2 = r^2

(49) + 25) = r^2

74 = r^2

Take the square root on both sides of the equation to find r.

sqrt{74} = sqrt{r^2}

sqrt{74} = r

It is ok to simplify the sqrt{74} but not needed.

We now have the three sides of the triangle that is form in quadrant 3.

We can now read cot θ from the triangle itself.

So, cot θ = adj/opp = (-7)/(-5) or 7/5.

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3 years ago
What is the measure of angle g?
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2 years ago
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                                                           Part A)

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y-y_1=m\left(x-x_1\right)

where

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substituting the values m = 6 and the point (x₁, y₁) = (7, 2) in the point-slope form of the line equation

y-y_1=m\left(x-x_1\right)

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                                                        Part B)

Given

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Using the point-slope form of the line equation

y-y_1=m\left(x-x_1\right)

where

m is the slope of the line

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As by theorem,

angle in a semicircle is a right angle. Angles that are in the same segment are equal.

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Learn more about circles here:

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\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-2}{1-(-4)}\implies \cfrac{10}{1+4}\implies \cfrac{10}{5}\implies 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=2[x-(-4)]\implies y-2=2(x+4) \\\\\\ y-2=2x+8\implies y=2x+10

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