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Mariana [72]
2 years ago
5

Which transformations can be used to map a triangle with vertices A (2, 2), B (4, 1), C (4, 5) to A’ (–2, –2), B’ (–1, –4), C’ (

–5, –4)?
a
a 90 degree clockwise rotation about the origin and a reflection over the y-axis

b
a 90 degree counterclockwise rotation about the origin and a translation down 4 units

c
a 180 degree rotation about the origin

d
a reflection over the y-axis and then a 90 degree clockwise rotation about the origin
Mathematics
1 answer:
Aliun [14]2 years ago
6 0
A is the the answer for you
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I NEED HELP . I'm having trouble trying to figure out how to answer this can anyone help please?
klio [65]

Answer:

what is it?

Step-by-step explanation:

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2 years ago
A scientist who studies teenage behavior was interested in determining if teenagers spend more time playing computer games then
Marina86 [1]

Answer:

- The amount of time greater than 10.2 hours for this year.

- The P_{value} is 0.01358.

Step-by-step explanation:

Given information:

mean, X = 10.2 hours

sample, n = 15

using t distribution

test statistics = 2.822

P_{value} = 2 x P(t_{n-1} > test statistics)

                              = 2 x P(t_{15-1} > 2.822)

                              = 2 x 0.00679

                              = 0.01358

P_{value} is less than 0.10 which indicates that the amount of time greater than 10.2 hours for this year.

5 0
2 years ago
The tables represent two linear functions in a system.<br><br> What is the solution to this system?
Darya [45]

The solution to this system is (x, y) = (8, -22).

The y-values get closer together by 2 units for each 2-unit increase in x. The difference at x=2 is 6, so we expect the difference in y-values to be zero when we increase x by 6 (from 2 to 8).

You can extend each table after the same pattern.

In table 1, x-values increase by 2 and y-values decrease by 8.

In table 2, x-values increase by 2 and y-values decrease by 6.

The attachment shows the tables extended to x=10. We note that the y-values are the same (-22) for x=8 (as we predicted above). That means the solution is ...

... (x, y) = (8, -22)

8 0
2 years ago
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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
2 years ago
The population of a city is 1,503,049. The land area of the city is 181.5 km².
Tresset [83]
1.) 8,281 people/km²
2.)
3.) 1226.88 grams
4.) 999.74 grams
5.) 6

I am not sure about number 2. 
5 0
2 years ago
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