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Stolb23
9 days ago
13

Xander says that it is not possible to draw a quadrilateral that is not a parallelogram and not a trapezoid. Devin says it is po

ssible. Explain who is correct.
A
Xander is correct because every trapezoid is a quadrilateral and a parallelogram.

B
Devin is correct because he can draw a quadrilateral with no pairs of parallel sides.

C
Xander is correct because both parallelograms and trapezoids have right angles.

D
Devin is right because parallelograms and trapezoids do not have the same number of sides.

Mathematics
1 answer:
Zina [12K]9 days ago
6 0
To address this question effectively, it's essential to clarify the characteristics of the given geometric terms.

A quadrilateral is defined as a four-sided figure.

Parallelograms are a subset of quadrilaterals characterized by having two pairs of parallel sides, with congruent opposite sides and angles. Instances of parallelograms include squares, rectangles, and rhombuses.

Trapezoids are also categorized under quadrilaterals but are distinguished by having only one pair of parallel sides.

In our investigation, we seek to determine whether there exists a four-sided shape without parallel sides. The included image presents an example of a quadrilateral that does not qualify as either a parallelogram or a trapezoid. Therefore, in accordance with the definitions of these polygons, the correct choice is B.

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Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f(x) is a real n
AnnZ [11958]

Answer:

Refer below

Step-by-step breakdown:

The function behaves as a piecewise function defined as:

N(t)=\left \{ {{25t+150} \ 0 \leq t \leq 6 \atop {\frac{200+80t}{2+0.05t} \ t \geq 8}} \right.

a)

We need to evaluate the limit of the function as t approaches infinity. This means determining the maximum number of fish present in the pond as time extends indefinitely.

We consider the second segment of the equation since t fits into that range, whereby t is infinite, clearly exceeding 8.

\lim_{t \to \infty} \frac{200+80t}{2+0.05t} \\\lim_{n \to \infty} \frac{80t}{0.05t}\\ \lim_{n \to \infty} \frac{80}{0.05} =1600

This indicates that the maximum fish population in this pond is 1600, regardless of the time.

b)

A function is considered continuous at a specific point if the limit and the function value at that point are the same.

The function value at t = 8, according to the second part of the equation, is:

\frac{200+80t}{2+0.05t}\\\frac{200+80(8)}{2+0.05(8)}\\=350

We observe that a value exists, and the limit approaches this as t nears 8.

<pthus>the function maintains continuity at t = 8

c)

We seek to determine if there exists a "time" during t from 0 to 6 when the fish count in the pond reaches 250. Substituting 250 into N(t) allows us to solve for t using the first portion of the piecewise function as shown below:

N(t)=25t+150\\250=25t+150\\25t=250-150\\25t=100\\t=4

The time is 4 years when the fish count in the pond becomes 250

</pthus>
6 0
1 month ago
An equation is written to represent the relationship between the temperature in Alaska during a snow storm, y, as it relates to
Leona [12162]
The options are: • only quadrant 1 • only quadrants 1 and 4. The reasoning is that the time since the storm began is always positive, implying positive values for x in quadrants 1 and 4. It’s noteworthy that during a blizzard, temperatures can be warmer than expected; some of the coldest snowstorms have temperatures between +5 °F and +18 °F. These numbers are below zero in Celsius, hence the corresponding quadrants depend on the temperature scale used. While Alaska often experiences temperatures well below freezing in either scale, snowfall typically requires the temperature to reach the levels mentioned before occurring. US temperatures are mostly reported in Fahrenheit, while historical data often records in Celsius. I lean towards "Quadrant 1 and 4 only", but one could argue for "1 only" or "4 only" as well.
8 0
25 days ago
The equation T^2 = A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun,
PIT_PIT [11945]

Let's denote the orbital period of planet X as T and its average distance from the sun as A. For planet Y, let its orbital period be T_1, implying that if planet Y's mean distance from the sun is twice that of planet X:

Z^2=(2A)^3\\ Z^2=8A^3\implies\\ Z^2=8T^2 \implies\\ Z=2\sqrt{2}T

This indicates that the orbital period for planet Y increases by a factor of 2\sqrt{2}

4 0
1 month ago
If (a^3+27)=(a+3)(a^2+ma+9) then m equals
Inessa [12191]

Answer:

m = - 3

Step-by-step explanation:

a³ + 27 can be recognized as a sum of cubes, which factors generally as

a³ + b³ = (a + b)(a² - ab + b²). Therefore:

a³ + 27

= a³ + 3³

= (a + 3)(a² - 3a + 9).

By comparing a² - 3a + 9 to a² + ma + 9, we find that

m = - 3.

7 0
1 month ago
ΔABC is an equilateral triangle. m∠A = (3x - 12)°. Solve for x.
zzz [11880]
In the case of an equilateral triangle ABC, each angle measures 60 degrees, as the total angle sum in any triangle is 180 degrees, and dividing that by 3 yields 60. Setting this equal to 3x-12, we have 60 = 3x - 12. After adding 12 to both sides, we get 72 = 3x, and dividing 72 by 3 gives us x = 24.66! I hope this clarifies things for you!
6 0
1 month ago
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