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3 years ago

Which rigid motion would map triangle ABC onto triangle A'B'C'?

Mathematics

2 answers:

KiRa [710]3 years ago

8 0

Answer:

It is either 3 or 4.

Step-by-step explanation:

Gelneren [198K]3 years ago

3 0

It would be choice 1 since when rotated all of the points would match up.

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A number cube is rolled and a coin is tossed. the number cube and the coin are fair. what is the probability that the number rol

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Suppose you roll a red number cube and a blue number cube. What is the probability that you will roll a 5 on the red cube and a 1 or 2 on the blue cube?

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3 years ago

I’m trying to get the answers to something just ignore this-

AURORKA [14]

Answer:

What is your question btw hello

Step-by-step explanation:

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2 years ago

PLEASE HELP ASAP. WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!

monitta

Answer:

the area of a regular hexagon is A=(3√3/2)a², a is the side length.

so A=(3√3/2)*5.2²=70.25

C is the correct answer

Step-by-step explanation: Hope this helps:)

6 0

3 years ago

Read 2 more answers

The table shows the volume of water in a tank after it has been filled to a certain height.

Lilit [14]

Functions and tables can be used to model real world situations.

The equation that represents volume is: $\mathbf{V(h) = 1.05(h)^3}$

From the table, we have:

$\mathbf{h \to V}$

$\mathbf{0 \to 0}$

$\mathbf{1 \to 1.05}$

$\mathbf{2 \to 8.40}$

$\mathbf{3 \to 28.35}$

The table can be rewritten as:

$\mathbf{h \to V}$

$\mathbf{0 \to 1.05 \times 0^3}$

$\mathbf{1 \to 1.05 \times 1^3}$

$\mathbf{2 \to 1.05 \times 2^3}$

$\mathbf{3 \to 1.05 \times 3^3}$

Replace 0, 1, 2 and 3 with h

$\mathbf{h \to 1.05 \times h^3}$

The above expression means that:

$\mathbf{V(h) = 1.05 \times h^3}$

Rewrite as:

$\mathbf{V(h) = 1.05(h)^3}$

Hence, the equation that represents volume is: $\mathbf{V(h) = 1.05(h)^3}$

Read more about equations and tables at:

brainly.com/question/19221147

5 0

3 years ago

A tank contains 60 kg of salt and 1000 L of water. A solution of a concentration 0.03 kg of salt per liter enters a tank at the

charle [14.2K]

Answer:

$\dfrac{dy}{dt}=0.27-0.009y(t),$ y(0)=60kg$

Step-by-step explanation:

Volume of water in the tank = 1000 L

Let y(t) denote the amount of salt in the tank at any time t.

Initially, the tank contains 60 kg of salt, therefore:

y(0)=60 kg

Rate In

A solution of concentration 0.03 kg of salt per liter enters a tank at the rate 9 L/min.

$R_{in}$ =(concentration of salt in inflow)(input rate of solution)

$=(0.03\frac{kg}{liter})( 9\frac{liter}{min})=0.27\frac{kg}{min}$

Rate Out

The solution is mixed and drains from the tank at the same rate.

Concentration, $C(t)=\dfrac{Amount}{Volume} =\dfrac{y(t)}{1000}$

$R_{out}$ =(concentration of salt in outflow)(output rate of solution)

$=\dfrac{y(t)}{1000}* 9\dfrac{liter}{min}=0.009y(t)\dfrac{kg}{min}$

Therefore, the differential equation for the amount of Salt in the Tank at any time t:

$\dfrac{dy}{dt}=R_{in}-R_{out}\\\\\dfrac{dy}{dt}=0.27-0.009y(t),$ y(0)=60kg$

8 0

4 years ago