All categories

3 years ago

Joe likes to go fishing. He knows that the stronger the fishing line is, the larger the fish that he can catch.

Mathematics

1 answer:

Trava [24]3 years ago

6 0

The answer is
B.

The size of the fish Joe can catch is a function of the strength of his fishing line.

You might be interested in

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

3 years ago

Use properties to find the sum or product 8 x 51

Lostsunrise [7]

8(51)

=8(50+1)

=8*50+8*1

=8*5*10+8*1

=400+8

=408

5 0

3 years ago

Help me out please! everyday !

Lady bird [3.3K]

Answer:

V = 904.32 ft^3

Step-by-step explanation:

The volume of a sphere is given by

V = 4/3 pi r^3

The radius is 6 and pi = 3.14

V = 4/3 ( 3.14) (6)^3

V = 904.31999 ft^3

Rounding to the hundredths place

V = 904.32 ft^3

5 0

3 years ago

Read 2 more answers

olga nikolaevna [1]

Answer:5

Step-by-step explanation:>

Which exponential function is represented by the

graph?

8Hold on, our servers are swamped. Wait for your answer to fully load.

o f(x) = 2( 3 )

O f(x) = {(2Y

o f(x) = {(

(1.1

61,0.25) 17

7605

Of(x) = 2(2)

-5 4 3 2 1

1 2 3 4

6 0

2 years ago

Determine whether the statement is always, sometimes, or never true. Complete the explanation. In parallelogram MNPQ, the diagon

Crank

This is never true. In a parallelogram, the diagonals will always bisect each other. Thus, each of these segments would always have to be equal.

4 0

3 years ago