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Natalka [10]
3 years ago
7

The endpoints of a side of rectangle ABCD in the coordinate plane are at A (2, 11) and B (7, 1). Find the equation of the line t

hat contains the given segment. The line segment is AD. The equation is y = .

Mathematics
2 answers:
Fittoniya [83]3 years ago
8 0

Answer:

Step-by-step explanation:

Given that ABCD is a rectangle. A is (2,11) and B (7,1)

To find equation of AD

AD is perpendicular to AB and passes through A

To find slope of AB

Slope = change in y/change in x = \frac{1-11}{7-2} =-2

Slope of AB = slope of perpendicular line = \frac{-1}{-2} =0.5

Using point slope formula we get equation of AD is

y-11=0.5(x-2)\\2y-22 =x-2\\x-2y+20 =0

xxTIMURxx [149]3 years ago
4 0
We know that
<span>A (2, 11) and B (7, 1)

step 1
find the slope of a line AB
the slope m=(y2-y1)/(x2-x1)------> m=</span>(1-11)/(7-2)---> m=-10/5----> m=-2

step 2
find the equation of a line AD

we know that
the segment AB and the segment AD are perpendicular
so
m1*m2=-1
m2=-1/m1-----> m2=1/2

with the slope m2=1/2 and the point A(2,11)
y-y1=m*(x-x1)------> y-11=(1/2)*(x-2)----> y=(1/2)x-1+11
y=(1/2)x+10----> y=0.5x+10

the answer is 
the equation of a line AD is y=0.5x+10

see the attached figure

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Vanyuwa [196]

Answer:

(i) the other two sides are 6 and  6\sqrt{2}

(ii) the other two sides are   \frac{4}{3} and                                         \frac{8}{3}

Step-by-step explanation:

(i)  Sine: sin(θ) = Opposite ÷ Hypotenuse

    Cosine: cos(θ) = Adjacent  ÷ Hypotenuse

    Tangent: tan(θ) = Opposite ÷ Adjacent

Here adjacent side = 6

opposite side = d

angle = 45°

other angles are 90° and 45°

tan (45) = Opposite ÷ Adjacent

 1 = d ÷ 6

∴ d = 6 × 1 = 6

so opposite side = 6

Hypotenuse ² = opposite side ² + adjacent side²

                      =  6² + 6²

                      = 36 + 36

                       = 72

hypotenuse = \sqrt{72}

                     = 6\sqrt{2}

the other two sides are 6 and  6\sqrt{2}

(ii) here adjacent side = 4√3

angle = 30°

other angles are 90° and 60°

opposite side = d

tan ( 30) = opposite ÷ adjacent

 \frac{1}{\sqrt{3}} = d ÷ 4√3

\frac{1}{\sqrt{3}} = d × (\frac{\sqrt{3}}{4})

                       3 d = 4

therefore d = \frac{4}{3}

therefore opposite side = \frac{4}{3}

Hypotenuse ² = opposite side ² + adjacent side²

                        =( \frac{4}{3})² +( \frac{4}{\sqrt{3}})²

                        = \frac{64}{9}

therefore hypotenuse = \sqrt{\frac{64}{9}}

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the other two sides are   \frac{4}{3} and                                    \frac{8}{3}

7 0
4 years ago
What is this awnser? 2/5y=4
MA_775_DIABLO [31]

Answer:

\frac{2}{5}y=4\quad :\quad y=10

4 0
3 years ago
Read 2 more answers
In the figure below ABCD is a rectangle and DA and CB are radii of circles tangent externally, AB = 4 cm. Provide the answer in
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Answer:

a. Radius of DA - 2 cm.

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d. Perimeter of shaded region - also 8-4π

<em><u>Hope this helps!  </u></em>

3 0
3 years ago
4p + 1 = 10 what is the value of p?
fredd [130]

Answer:

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Step-by-step explanation:

4p + 1 = 10

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p = 9/4 or 2.25

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25/40= .625
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