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3 years ago

What is the area of the shaded area

Mathematics

2 answers:

sdas [7]3 years ago

7 0

Answer:

cant tell tell me later

Step-by-step explanation:

jarptica [38.1K]3 years ago

6 0

Okay so first you need to find the height of the triangle (pythagorean theorm) a^2+b^2=c^2 so here we're looking for b
2^2 + b^2= 4^2
4+ b^2= 16
-4 -4
b^2=12 square root both b=appx 3.4
now we have the height and the base
area formula for a triangle is 1/2bh
4*3.4= 13.6 since there are 3 shaded triangles multiply by 3
13.6*3= 40.8 so the total area shaded is 40.8 hape this helped have a great day!!!

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What are the fourth roots of 6+6√(3i) ?

Helen [10]

Answer:

Step-by-step explanation:

The genral form of a complex number in rectangular plane is expressed as z = x+iy

In polar coordinate, z =rcos ∅+irsin∅ where;

r is the modulus = √x²+y²

∅ is teh argument = arctan y/x

Given thr complex number z = 6+6√(3)i

r = √6²+(6√3)²

r = √36+108

r = √144

r = 12

∅ = arctan 6√3/6

∅ = arctan √3

∅ = 60°

In polar form, z = 12(cos60°+isin60°)

z = 12(cosπ/3+isinπ/3)

To get the fourth root of the equation, we will use the de moivres theorem; zⁿ = rⁿ(cosn∅+isinn∅)

z^1/4 = 12^1/4(cosπ/12+isinπ/12)

When n = 1;

z1 = 12^1/4(cosπ/3+isinn/3)

z1 = 12^1/4cis(π/3)

when n = 2;

z2 = 12^1/4(cos2π/3+isin2π/3)

z2 = 12^1/4cis(2π/3)

when n = 3;

z2 = 12^1/4(cosπ+isinπ)

z2 = 12^1/4cis(π)

when n = 4;

z2 = 12^1/4(cos4π/3+isin4π/3)

z2 = 12^1/4cis(4π/3)

8 0

3 years ago

What is the average rate of change for the function ( IN THE PIC) over the interval (also in pic) HELP PLSSS ASAP!!!!

solong [7]

It is B because I had the question before in my school

3 0

3 years ago

A ladder 16 feet long is leaning against the wall of a tall building. The base of the ladder is moving away from the wall at a r

svet-max [94.6K]

Answer:

a. 0.588

b. 0.0722

c. 4.576 sqft/sec

Step-by-step explanation:

Let b and h denote the base and height as indicated in the diagram. By pythagoras theorem, $h^2 + b^2 = 16^2 = 256 \dotsc\;(1)$ because it is a right angle triangle.

It is given that $\frac{db}{dt} = 1$

Now differentiate (1) with respect to t (time) :

$\displaystyle{2h\frac{dh}{dt} + 2b\frac{db}{dt} = 0 \implies \frac{dh}{dt} = -\frac{b}{h} \frac{db}{dt}}$

$\displaystyle{=-\frac{b}{\sqrt{256 - b^2}} \frac{db}{dt} = -\frac{8}{13.856} \times 1 = -0.588}$

The minus sign indicates that the value of h is actually decreasing. The required answer is 0.588.

b. From the diagram, infer that $16 \sin{\theta} = b$ . When b = 8, then $\theta = \arcsin{0.5} = \ang{30}$ .

Differentiate the above equation w.r.t t

$\displaystyle{16 \cos{\theta} \frac{d\theta}{dt} = \frac{db}{dt} \implies \frac{d\theta}{dt} = \frac{1}{16 \cos{\theta}} = \frac{1}{13.856} = \mathbf{0.0722}}$