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Luden [163]
3 years ago
5

Polygon ABCD has the following vertices:

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
8 0

9514 1404 393

Answer:

  (b)  38.5

Step-by-step explanation:

The polygon is a trapezoid with bases 4 and 7 and height 7. Its area is given by the formula ...

  A = (1/2)(b1 +b2)h

  A = (1/2)(4 +7)(7) = 77/2

  A = 38.5 . . . . square units

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Express 112 as a product of prime numbers in index form?
alexgriva [62]
                                   112
                                    /  \
                                  2   56
                                       /  \
                                     2   28
                                          /  \
                                         2  14
                                              /  \
                                             2  7

So, the answer is 2 x 2 x 2 x 7 or 2^3 x 7
6 0
3 years ago
Help! I’ll give extra points:)
Fofino [41]

Answer:

if I'm not wrong I think it's B

3 0
2 years ago
Help me please!!<br> hurry
zubka84 [21]

Answer:

n-3

Step-by-step explanation:

for every input, the output is three less (4-3=1, 7-3=4, 8-3=5). That rule needs to stay consistent, so no matter what the input is (n), the output is always going to be three less than, making it y or output=n-3

6 0
3 years ago
Write an equation that represents the line.<br> Use exact numbers.<br> (-2,-1)<br> (4,6)
vazorg [7]

Check the picture below.

to get the equation of any straight line, we simply need two points off of it, so let's use those in the picture

(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{6})~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{6}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{(-2)}}} \implies \cfrac{6 -1}{4 +2} \implies \cfrac{ 5 }{ 6 }

\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-\stackrel{y_1}{1}=\stackrel{m}{\cfrac{5}{6}}(x-\stackrel{x_1}{(-2)})\implies y-1=\cfrac{5}{6}(x+2) \\\\\\ y-1=\cfrac{5}{6}x+\cfrac{5}{3}\implies y=\cfrac{5}{6}x+\cfrac{5}{3}+1\implies y=\cfrac{5}{6}x+\cfrac{8}{3}

6 0
2 years ago
QUESTION 1
Kitty [74]

It's difficult to make out what the force and displacement vectors are supposed to be, so I'll generalize.

Let <em>θ</em> be the angle between the force vector <em>F</em> and the displacement vector <em>r</em>. The work <em>W</em> done by <em>F</em> in the direction of <em>r</em> is

<em>W</em> = <em>F</em> • <em>r</em> cos(<em>θ</em>)

The cosine of the angle between the vectors can be obtained from the dot product identity,

<em>a</em> • <em>b</em> = ||<em>a</em>|| ||<em>b</em>|| cos(<em>θ</em>)   ==>   cos(<em>θ</em>) = (<em>a</em> • <em>b</em>) / (||<em>a</em>|| ||<em>b</em>||)

so that

<em>W</em> = (<em>F</em> • <em>r</em>)² / (||<em>F</em>|| ||<em>r</em>||)

For instance, if <em>F</em> = 3<em>i</em> + <em>j</em> + <em>k</em> and <em>r</em> = 7<em>i</em> - 7<em>j</em> - <em>k</em> (which is my closest guess to the given vectors' components), then the work done by <em>F</em> along <em>r</em> is

<em>W</em> = ((3<em>i</em> + <em>j</em> + <em>k</em>) • (7<em>i</em> - 7<em>j</em> - <em>k</em>))² / (√(3² + 1² + 1²) √(7² + (-7)² + (-1)²))

==>   <em>W</em> ≈ 5.12 J

(assuming <em>F</em> and <em>r</em> are measured in Newtons (N) and meters (m), respectively).

3 0
3 years ago
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