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3 years ago

Match each condition to the number of quadrilaterals that can be constructed to fit the condition.

Mathematics

2 answers:

frozen [14]3 years ago

8 0

Answer:

a, c, and d go with many quadrilaterals. b, e, and f, go with one quadrilateral

Step-by-step explanation:

Sedbober [7]3 years ago

5 0

Answer:

See the explanation.

Step-by-step explanation:

Condition A.

A rectangle with four right angles

There can be many quadrilaterals satisfying this condition.

Condition B.

A square with one side measuring 5 inches

There can be only one quadrilateral satisfying this condition.

Condition C.

A rhombus with one angle measuring 43°

There can be many quadrilaterals satisfying this condition.

Condition D.

A parallelogram with one angle measuring 32°

There can be many quadrilaterals satisfying this condition.

Condition E.

A parallelogram with one angle measuring 48° and adjacent sides measuring 6 inches and 8 inches.

There can be only one quadrilateral satisfying this condition.

Condition F.

A rectangle with adjacent sides measuring 4 inches and 3 inches.

There can be only one quadrilateral satisfying this condition.

(Answer)

You might be interested in

When a distribution is mound-shaped symmetrical, what is the general relationship among the values of the mean, median, and mode

yuradex [85]

Answer:

The mean, median, and mode are approximately equal.

Step-by-step explanation:

The mean, median, and mode are central tendency measures in a distribution. That is, they are measures that correspond to a value that represents, roughly speaking, "the center" of the data distribution.

In the case of a normal distribution, these measures are located at the same point (i.e., mean = median = mode) and the values for this type of distribution are symmetrically distributed above and below the mean (mean = median = mode).

When a distribution is not symmetrical, we say it is skewed. The skewness is a measure of the asymmetry of the distribution. In this case, the mean, median and mode are not the same, and we have different possibilities as the mentioned in the question: the mean is less than the median and the mode (negative skew), or greater than them (positive skew), or approximately equal than the median but much greater than the mode (a variation of a positive skew case).

In the case of the normal distribution, the skewness is 0 (zero).

Therefore, in the case of a mound-shaped symmetrical distribution, it resembles the normal distribution and, as a result, it has similar characteristics for the mean, the median, and the mode, that is, they are all approximately equal. So, the general relationship among the values for these central tendency measures is that they are all approximately equal for mound-shaped symmetrical distributions, considering they have similar characteristics of the normal distribution, which is also a mound-shaped symmetrical distribution (as well as the t-student distribution).

5 0

2 years ago

If you are offered one slice from a round pizza (in other words, a sector of a circle) and the slice must have a perimeter of 24

VARVARA [1.3K]

Answer:

A 12 in diameter will reward you with the largest slice of pizza.

Step-by-step explanation:

Let $r$ be the radius and $\theta$ be the angle of a circle.

According with the graph, the area of the sector is given by

$A=\frac{1}{2}r^2\theta$

The arc lenght of a circle with radius $r$ and angle $\theta$ is $r \theta$

The perimeter of the pizza slice is composed of two straight pieces, each of length r inches, and an arc of the circle which you know has length s = rθ inches. Thus the perimeter has length

$2r+r\theta=24 \:in$

We need to express the area as a function of one variable, to do this we use the above equation and we solve for $\theta$

$2r+r\theta=24\\r\theta=24-2r\\\theta=\frac{24-2r}{r}$

Next, we substitute this equation into the area equation

$A=\frac{1}{2}r^2(\frac{24-2r}{r})\\A=\frac{1}{2}r(24-2r)\\A=12r-r^2$

The domain of the area is

$0$

To find the diameter of pizza that will reward you with the largest slice you need to find the derivative of the area and set it equal to zero to find the critical points.

$\frac{d}{dr} A=\frac{d}{dr}(12r-r^2)\\A'(r)=\frac{d}{dr}(12r)-\frac{d}{dr}(r^2)\\A'(r)=12-2r$

$12-2r=0\\-2r=-12\\\frac{-2r}{-2}=\frac{-12}{-2}\\r=6$

To check if $r=6$ is a maximum we use the Second Derivative test

if $f'(c)=0$ and $f''(c)$ , then f(x) has a local maximum at x = c.

The second derivative is

$\frac{d}{dr} A'(r)=\frac{d}{dr} (12-2r)\\A''(r)=-2$

Because $A''(r)=-2$ the largest slice is when r = 6 in.

The diameter of the pizza is given by

$D=2r=2\cdot 6=12 \:in$

A 12 in diameter will reward you with the largest slice of pizza.

3 0

2 years ago

Which represents the solution to the absolute value equation? 3|2x+4|-1=11

larisa86 [58]

The first step for absolute value equations is to isolate the expression contained within the absolute value bars:

3|2x+4|-1 = 11
3|2x+4| = 12
|2x+4| = 4

so |2x+4| is 4 units away from 0 on a number line, but we don't know in which direction -- negative or positive? you'll have two answers.

2x+4 = 4

AND

2x+4 = -4

solve both of those two step equations and you'll get

x = 0

AND

x = -4

so 0 and -4 are your solutions.

5 0

3 years ago

Sis Avila (STUDEN...

vesna_86 [32]

Answer:

sorry I dont know the way u put it makes no sense

3 0

3 years ago

I need help with this question!

Bumek [7]

A)
A polygon is a closed shape so there must be at least three sides. Therefore, the least value that n can be would be three.

b)
n value
3 0
n(n-3)/2
3(3-3)/3
3(0)/3
0/3
0

4 2
n(n-3)/2
4(4-3)/2
4(1)/2
4/2
2

5 5
n(n-3)/2
5(5-3)/2
5(2)/2
10/2
5

6 0

3 years ago

Match each condition to the number of quadrilaterals that can be constructed to fit the condition.​

Match each condition to the number of quadrilaterals that can be constructed to fit the condition.