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2 years ago

Explain the tangent line problem

1 answer:

8 0

The Tangent Line Problem 1/3How do you find the slope of the tangent line to a function at a point Q when you only have that one point? This Demonstration shows that a secant line can be used to approximate the tangent line. The secant line PQ connects the point of tangency to another point P on the graph of the function. As the distance between the two points decreases, the secant line becomes closer to the tangent line.

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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

6 0

2 years ago

A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with A random s

wel

Answer:

95%: (3278.354 ; 3270.083)

99% : (3221.646 ; 3278.354)

Step-by-step explanation:

Given :

Sample size, n = 12

Mean, xbar = 3250

Sample standard deviation = √1000

The 95% confidence interval :

Mean ± Margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 0.05, df=12-1 = 11 ;

Tcritical at 95% = 2.20

Hence,

Margin of Error = (2.20 * √1000/√12) = 20.083

Confidence interval : 3250 ± 20.083

Lower boundary = 3250 - 20.083 = 3229.917

Upper boundary = 3250 + 20.083 = 3270.083

2.)

The 99% confidence interval :

Mean ± Margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 0.01, df=12-1 = 11 ;

Tcritical at 99% = 3.106

Hence,

Margin of Error = (3.106 * √1000/√12) = 28.354

Confidence interval : 3250 ± 28.354

Lower boundary = 3250 - 28.354 = 3221.646

Upper boundary = 3250 + 28.354 = 3278.354

3 0

3 years ago

Is this equation an infinite solution, no solution, or one solution?? 7(2+5x) = 3x + 14

shusha [124]

Answer:

One Solution

Step-by-step explanation:

Once you solve the equation, you get 0 which satisfies for X, therefore it has a solution.

7 0

2 years ago

Lines l and m are parallel. Use the diagram to determine the measure of ∠3.

Pavlova-9 [17]

Answer:

∠3 is 70°

Step-by-step explanation:

∠1 is 50° since it is across from another 50° angle

and then you use the equation x+25+2x+50=180 to find x

x then equals 35°

you then use the alternate interior angle theorem to say that ∠3 is 2x, or 70°

5 0

2 years ago

If f(x) = 3x - 1 and g(x) = x + 2, find (f+g)(x).

yawa3891 [41]

F+g(x) = f(x) + g(x) = 3x-1 + x+2 = 4x+1

5 0

2 years ago