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Galina-37 [17]
3 years ago
14

What is the order of rotational symmetry for the figure? A. 3 B. 1 C. 2 D. 4 or more

Mathematics
2 answers:
Anastasy [175]3 years ago
4 0

Answer:

This rotational symmetry has 4 or more order .

OAmalOHopeO

Advocard [28]3 years ago
4 0

9514 1404 393

Answer:

  B.  1

Step-by-step explanation:

The figure has no symmetry, so only maps to itself with 360° of rotation.

The order of rotational symmetry is 1.

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Simple question just answer correctly to get brainleist and 10 pts! :)
Natali [406]

Answer:

d.  The mapping represents y as a function of x, because each x-value corresponds to exactly one y-value.

Step-by-step explanation:

This would be a function as long as one x-value doesn't correspond to two y-values. If two x-values correspond to one y-value though, it still represents a function.

8 0
3 years ago
Read 2 more answers
A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k
Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0
3 years ago
10
schepotkina [342]

Answer:

3.56 cm.

Step-by-step explanation:

1/3 π r^2 h = V

1/3 π (5.13)^2 h  = 98

h = 98 / (1/3 π (5.13)^2)

=  98 / 27.55899

= 3.556.

3 0
3 years ago
URGENT! PLS ANSWER QUICK!!!!!!!!
skad [1K]

Since the lines have the same slopes, hence they are parallel lines

<h3>How to determine the relationship between lines</h3>

We can determine the relations by knowing the slopes of the line

For the line with coordinates (9, –5) and (5, –2)

Slope = -2+5/5-9
Slope = -3/4

For the line with coordinates (–4, –2) and (–8, 1)

Slope = 1+2/-8+4
Slope = -3/4

Since the lines have the same slopes, hence they are parallel lines

Learn more on parallel lines here: brainly.com/question/16742265

#SPJ1

7 0
2 years ago
How can Doug use place value pattern to check his answer?
MAVERICK [17]
He can use it to make sure he lined his place values up properly
3 0
3 years ago
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