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Damm [24]
3 years ago
5

Which of the following statements best represents the relationship between a relation and a function. a function is always a rel

ation but a relation is not always a function. a relation is always a function but a function is not always a relation. a relation is never a function but a function is always a relation.?
Mathematics
2 answers:
hoa [83]3 years ago
8 0

Answer:

A function is always a relation but a relation is not always a function

Step-by-step explanation:

In mathematics the relation shows the relationship of the elements of a set to the elements of another set,

For example,

If A = { a, b, c }, B = {e, f, g, h}

R = { (a, e), (a, f), (b, g), (c, h)}

Then R is called relation from A to B,

Where, A is called Domain and B is called co-domain.

A relation is called function if the elements of domain are related to only one element of co domain.

i.e.

If R = { (a, e), (b, f), (c, g) }

Then R is called function.

Hence, the correct option is,

A function is always a relation but a relation is not always a function

aleksandrvk [35]3 years ago
3 0
The first statement: a function is always a function but a relation is not always a function.
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gabys piggy bank contains nickles and dimes worth $5.15. If she has 66 coins in all, how many of each does she have
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8.18

Step-by-step explanation:

8.18

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Quadrilateral ABCD has vertices A (-3, 0), B (2, 4), C (3, 1), and D (-4, -3). Calculate the perimeter of the quadrilateral. Rou
Andru [333]

Answer:

Perimeter of the Quadrilateral=20.78

Step-by-step explanation:

Perimeter of the Quadrilateral= AB+BC+CD+AD

Finding the sides of the quadrilateral using the Distance formula:

A(-3,0), B(2,4), C(3,1) and D(-4,-3)

AB=\sqrt{(2-(-3))^2+(4-0)^2}\\\\ =\sqrt{(2+3)^2+4^2}\\\\ =\sqrt{5^2+4^2} \\\\=\sqrt{25+16}\\\\ =\sqrt{41}\\

=6.40

BC=

         \sqrt{(3-2)^2+(1-4)^2} \\\\=\sqrt{1^2+(-3)^2} \\\\=\sqrt{1+9}\\\\ =\sqrt{10}

          =3.16

      CD=\sqrt{(-4-3)^2+(-3-1)^2} \\\\=\sqrt{(-7)^2+(-4)^2} \\\\=\sqrt{49+16}\\\\=\sqrt{65}\\\\ =8.06

AD=\sqrt{(-4-(-3))^2+(-3-0)^2} \\\\=\sqrt{(-4+3)^2+(-3)^2} \\\\=\sqrt{(-1)^2+(-3)^2} \\\\=\sqrt{1+9}\\\\=\sqrt{10} \\\\ =3.16

Perimeter of the quadrilateral= AB+BC+CD+AD

             =6.40+3.16+8.06+3.16

              =20.78

6 0
3 years ago
A conical paper funnel of radius 4 and height 6 units is needed to make a good cup of coffee. If this funnel is made out of a se
Nimfa-mama [501]

Answer:

The radius of this circle is 2\sqrt{13} and the central angle of the sector is \theta=\frac{2}{\sqrt{13}}\times360^\circ.

Step-by-step explanation:

Consider the provided information.

If a conical paper funnel is made out of a sector of circle then the radius of the sector becomes the slant height of the cone, and the length of the curved part of the sector becomes the circumference of the base of the cone.

First find the slant height: l=\sqrt{r^2+h^2}

l=\sqrt{4^2+6^2}

l=\sqrt{16+36}

l=2\sqrt{13}

Thus, the radius of the sector is 2\sqrt{13} which is same as the slant height of the cone.

Now, the circumference of the base of the cone is same as the length of the curved part.

C=2\pi r

C=2\pi \times4=8\pi

The length of the curved part = \frac{\theta}{360\circ}\times2\pi r

\frac{\theta}{360\circ}\times2\pi \times2\sqrt{13}=8\pi

\frac{\theta}{360\circ}\times\sqrt{13}=2

\theta=\frac{2}{\sqrt{13}}\times360^\circ

Hence, the radius of this circle is 2\sqrt{13} and the central angle of the sector is \theta=\frac{2}{\sqrt{13}}\times360^\circ.

4 0
4 years ago
the mean weight of 32 math students is 43.5. if the students could all stand on a scale together, what would their total weight
Lera25 [3.4K]
1392
Sum of weight=mean•number of people
1392=43.5•32
7 0
4 years ago
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