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Rashid [163]
3 years ago
8

Drag each function to the correct location on the chart.

Mathematics
1 answer:
Oliga [24]3 years ago
8 0
It’s even I just did the test
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PLEASE HELP!!!
maxonik [38]

people cant read that

5 0
3 years ago
Write an expression for the calculation double 2 and then add 5.
Mariana [72]

The expression is (2x2)+5

8 0
3 years ago
How many packages of hamburger buns and meat patties should you buy to sell at a baseball tournament?
Mandarinka [93]
13 packs of meat patties and 16 packs of buns
4 0
4 years ago
(a) Let R = {(a,b): a² + 3b <= 12, a, b € z+} be a relation defined on z+)
grin007 [14]

Answer:

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Step-by-step explanation:

The relation R is an equivalence if it is reflexive, symmetric and transitive.

The order to options required to show that R is an equivalence relation are;

((a, b), (a, b)) ∈ R since a·b = b·a

Therefore, R is reflexive

If ((a, b), (c, d)) ∈ R then a·d = b·c, which gives c·b = d·a, then ((c, d), (a, b)) ∈ R

Therefore, R is symmetric

If ((c, d), (e, f)) ∈ R, and ((a, b), (c, d)) ∈ R therefore, c·f = d·e, and a·d = b·c

Multiplying gives, a·f·c·d = b·e·c·d, which gives, a·f = b·e, then ((a, b), (e, f)) ∈R

Therefore R is transitive

From the above proofs, the relation R is reflexive, symmetric, and transitive, therefore, R is an equivalent relation.

Reasons:

Prove that the relation R is reflexive

Reflexive property is a property is the property that a number has a value that it posses (it is equal to itself)

The given relation is ((a, b), (c, d)) ∈ R if and only if a·d = b·c

By multiplication property of equality; a·b = b·a

Therefore;

((a, b), (a, b)) ∈ R

The relation, R, is reflexive.

Prove that the relation, R, is symmetric

Given that if ((a, b), (c, d)) ∈ R then we have, a·d = b·c

Therefore, c·b = d·a implies ((c, d), (a, b)) ∈ R

((a, b), (c, d)) and ((c, d), (a, b)) are symmetric.

Therefore, the relation, R, is symmetric.

Prove that R is transitive

Symbolically, transitive property is as follows; If x = y, and y = z, then x = z

From the given relation, ((a, b), (c, d)) ∈ R, then a·d = b·c

Therefore, ((c, d), (e, f)) ∈ R, then c·f = d·e

By multiplication, a·d × c·f = b·c × d·e

a·d·c·f = b·c·d·e

Therefore;

a·f·c·d = b·e·c·d

a·f = b·e

Which gives;

((a, b), (e, f)) ∈ R, therefore, the relation, R, is transitive.

Therefore;

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Based on a similar question posted online, it is required to rank the given options in the order to show that R is an equivalence relation.

Learn more about equivalent relations here:

brainly.com/question/1503196

4 0
3 years ago
What's the surface area ratio &amp; the volume ratio??<br><br> please help me ASAP!!
Ulleksa [173]

Answer:

Step-by-step explanation:

Volumes of two spheres A and B = 648 cm³ and 1029 cm³

Things to remember:

1). Scale factor of two objects = \frac{r_1}{r_2} [r_1 and r_2 are the radii of two circles]

2). Area scale factor = \frac{(r_1)^2}{(r_2)^2}

3). Volume scale factor = \frac{(r_1)^3}{(r_2)^3}

Volume scale factor Or Volume ratio = \frac{V_A}{V_B}

                         \frac{(r_1)^3}{(r_2)^3}= \frac{648}{1029}

                         \frac{r_1}{r_2}=\sqrt[3]{\frac{648}{1029} }

                         \frac{r_1}{r_2}=\frac{6(\sqrt[3]{3})}{7(\sqrt[3]{3})}

                        \frac{r_1}{r_2}=\frac{6}{7}

Therefore, scale factor = \frac{r_1}{r_2}=\frac{6}{7}

                                      ≈ 6 : 7

Area scale factor Or area ratio = (\frac{r_1}{r_2})^2=(\frac{6}{7})^2

                                                   = \frac{36}{49}

                                                   ≈ 36 : 49

Volume scale factor or Volume ratio = \frac{648}{1029}

                                                             = \frac{216}{343}

                                                             ≈ 216 : 343

4 0
3 years ago
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