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3 years ago

Find the angle between the following pairs of lines x^2+6xy +9y^2-4x +12y-5 =0

Mathematics

1 answer:

wlad13 [49]3 years ago

3 0

Answer:

+6xy+9y

+4x+12y−5=0

Step-by-step explanation:

+6xy+9y

+4x+12y−5=0

Comparing the equation with the general equation of second degree gives

a=1,b=9,h=3,g=2,f=6,c=−5

Angle between a pair of straight lines that is tanθ=

∣

a+b

−ab

∣

tanθ=

∣

1+9

9−1×9

∣

tanθ=0

⇒θ=tan

−

(0)=0

∘

Angle between the pair of straight lines is zero therefore the lines are parallel.

Hence proved

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A garden is shaped in the form of a regular heptagon (seven-sided), MNSRQPO. A circle with center T and radius 25m circumscribes

Alenkinab [10]

The relationship between the sides MN, MS, and MQ in the given regular heptagon is $\dfrac{1}{MN} = \dfrac{1}{MS} + \dfrac{1}{MQ}$

The area to be planted with flowers is approximately 923.558 m²

The reason the above value is correct is as follows;

The known parameters of the garden are;

The radius of the circle that circumscribes the heptagon, r = 25 m

The area left for the children playground = ΔMSQ

Required;

The area of the garden planted with flowers

Solution:

The area of an heptagon, is;

$A = \dfrac{7}{4} \cdot a^2 \cdot cot \left (\dfrac{180 ^{\circ}}{7} \right )$

The interior angle of an heptagon = 128.571°

The length of a side, S, is given as follows;

$\dfrac{s}{sin(180 - 128.571)} = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)}$

$s = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)} \times sin(180 - 128.571) \approx 21.69$

$The \ apothem \ a = 25 \times sin \left ( \dfrac{128.571}{2} \right) \approx 22.52$

The area of the heptagon MNSRQPO is therefore;

$A = \dfrac{7}{4} \times 22.52^2 \times cot \left (\dfrac{180 ^{\circ}}{7} \right ) \approx 1,842.94$

$MS = \sqrt{(21.69^2 + 21.69^2 - 2 \times 21.69 \times21.69\times cos(128.571^{\circ})) \approx 43.08$

By sine rule, we have

$\dfrac{21.69}{sin(\angle NSM)} = \dfrac{43.08}{sin(128.571 ^{\circ})}$

$sin(\angle NSM) =\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ})$

$\angle NSM = arcsin \left(\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ}) \right) \approx 23.18^{\circ}$

∠MSQ = 128.571 - 2*23.18 = 82.211

The area of triangle, MSQ, is given as follows;

$Area \ of \Delta MSQ = \dfrac{1}{2} \times 43.08^2 \times sin(82.211^{\circ}) \approx 919.382^{\circ}$

The area of the of the garden plated with flowers, $A_{req}$ , is given as follows;

$A_{req}$ = Area of heptagon MNSRQPO - Area of triangle ΔMSQ

Therefore;

$A_{req}$ = 1,842.94 - 919.382 ≈ 923.558

The area of the of the garden plated with flowers, $A_{req}$ ≈ 923.558 m²

Learn more about figures circumscribed by a circle here:

brainly.com/question/16478185

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3 years ago

I need help on this question and answering it

Contact [7]

Answer:

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Step-by-step explanation:

the shadow to the object would be a 4:5 ratio

since the shadow of the flagpole is 60 ft,

4x=60 so x would =15

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A triangle has side lengths of 34 in., 20 in., and 47 in. Classify it as acute, obtuse, or right.

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Find the angle between the following pairs of lines x^2+6xy +9y^2-4x +12y-5 =0​

Find the angle between the following pairs of lines x^2+6xy +9y^2-4x +12y-5 =0