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yaroslaw [1]
3 years ago
8

Which set of shapes contains members that are always similar to one another? trapezoids isosceles triangles equilateral triangle

s rectangles pentagons
Mathematics
2 answers:
andrew11 [14]3 years ago
5 0

Answer: Equilateral triangles.


Step-by-step explanation:

We know that in similar polygons, its all the corresponding angles are congruent.

Only the set of equilateral triangles contains members that are always similar( by AAA similarity criteria ) to one another as the measure of all the angles of equilateral triangle is fixed i.e. 60°.

Rest other do not have fixed measure for angles in it.

alekssr [168]3 years ago
4 0
The answer is equilateral triangles. It says it in it's name that they will always be similar to one another.
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Answer:

  1. A rational function is any function which can be written as the ratio of two polynomial functions, where the polynomial in the denominator is not equal to zero. The domain of f(x)=P(x)Q(x) f ( x ) = P ( x ) Q ( x ) is the set of all points x for which the denominator Q(x) is not zero
  2. To determine the x-intercept, we set y equal to zero and solve for x. Similarly, to determine the y-intercept, we set x equal to zero and solve for y
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refer this attachment for 1st question ( given the rational function f(x)=2x+6/x-3, Answer the following questions. )

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3 years ago
When an amount of heat Q (in kcal) is added to a unit mass (kg) of a
lisov135 [29]

Answer:

The change in temperature per minute for the sample, dT/dt is 71.\overline {6} °C/min

Step-by-step explanation:

The given parameters of the question are;

The specific heat capacity for glass, dQ/dT = 0.18 (kcal/°C)

The heat transfer rate for 1 kg of glass at 20.0 °C, dQ/dt = 12.9 kcal/min

Given that both dQ/dT and dQ/dt are known, we have;

\dfrac{dQ}{dT} = 0.18 \, (kcal/ ^{\circ} C)

\dfrac{dQ}{dt} = 12.9 \, (kcal/ min)

Therefore, we get;

\dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = {\dfrac{dQ}{dt} } \times \dfrac{dT}{dQ} = \dfrac{dT}{dt}

\dfrac{dT}{dt} = \dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = \dfrac{12.9 \, kcal / min }{0.18 \, kcal/ ^{\circ} C }   = 71.\overline 6 \, ^{\circ } C/min

For the sample, we have the change in temperature per minute, dT/dt, presented as follows;

\dfrac{dT}{dt}  = 71.\overline 6 \, ^{\circ } C/min

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Step-by-step explanation:

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