The smallest spider , is 0.43 milimeters long , a scale of this spider is 8 centimeters long , what’s the scale of the model?

Equilateral triangle ABC has an area of \sqrt{3}√ 3 . If the shaded region has an area of \piπK − \sqrt{3}√ 3 , what

Liono4ka [1.6K]

Answer:

The value of k = 4/3

Step-by-step explanation:

* Lets explain how to solve the problem

- An equilateral triangle ABC is inscribed in a circle N

- The area of the triangle is √3

- The shaded area is the difference between the area of the circle

and the area of the equilateral triangle ABC

- The shaded are = k π - √3

- We need to find the value of k

* At first lets find the length of the side of the Δ ABC

∵ Δ ABC is an equilateral triangle

∴ Its area = √3/4 s² , where s is the length of its sides

∵ The area of the triangle = √3

∴ √3/4 s² = √3

- divide both sides by √3

∴ 1/4 s² = 1

- Multiply both sides by 4

∴ s² = 4 ⇒ take √ for both sides

∴ s = 2

∴ The length of the side of the equilateral triangle is 2

* Now lets find the radius of the circle

- In the triangle whose vertices are A , B and N the center of the circle

∵ AN and BN are radii

∴ AN = BN = r , where r is the radius of the circle

∵ The sides of the equilateral angles divides the circle into 3 equal

arcs in measure where each arc has measure 360°/3 = 120°

∵ The measure of the central angle in a circle equal the measure

of the its subtended arc arc

∵ ∠ANB is an central angle subtended by arc AB

∵ The measure of arc AB is 120°

∴ m∠ANB = 120°

- By using the cosine rule in Δ ANB

∵ AB = 2 , AN = BN = r , m∠ANB = 120°

∴ $(2)^{2}=r^{2}+r^{2}-2(r)(r)cos(120)$

∴ $4=r^{2}+r^{2}-2(r)(r)(-0.5)$

∴ $4=r^{2}+r^{2}-(-r^{2})$

∴ $4=r^{2}+r^{2}+r^{2}$

∴ $4=3r^{2}$

- Divide both sides by 3

∴ $r^{2}=\frac{4}{3}$

- Take square root for both sides

∴ r = 2/√3

* Lets find the value of k

∵ Area circle = πr²

∵ r = 2/√3

∴ Area circle = π(2/√3)² = (4/3)π

∵ Area shaded = area circle - area triangle

∵ Area triangle = √3

∴ Area shaded = (4/3) π - √3

∵ Area of the shaded part is π k - √3

- Equate the two expressions

∴ π k - √3 = (4/3) π - √3

∴ k = 4/3

* The value of k = 4/3

7 0

2 years ago

I feel hurt so bad for some reason

crimeas [40]

Like physical or emotional ?

4 0

3 years ago

Read 2 more answers

Akira buys 10 tickets to a show. She also pays a $5 parking fee. If she spent $35 in total, how much did each ticket cost? Use x

Arte-miy333 [17]

Answer: Each ticket cost 3 dollars

Step-by-step explanation: First take 35 and subtract 5 which equals 30

then divide 30 by 10 and you get 3

7 0

3 years ago

Drag each label to the correct location on the table. Each label can be used more than once, but not all labels will be used.

Natalka [10]

Answer:

what lable i dont see nothing

Step-by-step explanation:

7 0

2 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago