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Salsk061 [2.6K]
3 years ago
9

Is (3,13) a solution of y = 2x - y

Mathematics
1 answer:
PtichkaEL [24]3 years ago
5 0

Answer:

NO

Step-by-step explanation:

(x, y) so if you plug in your points... 13 = 2(3) - 13

To solve:

Multiply:    13 = 2(3) - 13

Subtract:   13 = 6 - 13

Answer:     13 does not equal -7

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Shira has 20 books on her bookshelf. Of those 15 are about horses, the rest are nature books. What fraction of Shiras books are
givi [52]

Answer:

1/4

Step-by-step explanation:

15 out of the 20 books are horses. 5 of those are nature. 5/20 is the fraction, but can be simplified to 1/4

6 0
3 years ago
The sum of five consecutive integers is 55. Find the integers.​
liubo4ka [24]

Answer:

9, 10, 11, 12, 13

Step-by-step explanation:

(x-2)+(x-1)+x+(x+1)+(x+2) = 55

5x = 55

x = 11

8 0
3 years ago
Which compares the modes of the data?
sveta [45]

Answer:

I think C

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
You are given a choice of taking the simple interest on ​$10,000 invested for 2 years at a rate of 3​% or the interest on ​$100,
attashe74 [19]

Answer:

Compount interest earns more. Difference between 2 interest is $92 445.39

Step-by-step explanation:

Simple Interest:

I =  \frac{prt}{100}

p = $10000

r = 3%

t = 2years

I = (10000×3×2)/100

= $600

Total amount = $10 600

Compound Interest:

A = p( {1 +  \frac{r}{100}) }^{n}

p = $100000

r = 3/730 (daily)

t = 730 (2yrs)

A = 100000[1+(3/73000)]^730

= $103 045.39 (2d.p)

Difference = $103045.39 -

$10600

= $92 445.39

(Correct me if i am wrong)

8 0
3 years ago
given examples of relations that have the following properties 1) relexive in some set A and symmetric but not transitive 2) equ
rodikova [14]

Answer: 1) R = {(a, a), (а,b), (b, a), (b, b), (с, с), (b, с), (с, b)}.

It is clearly not transitive since (a, b) ∈ R and (b, c) ∈ R whilst (a, c) ¢ R. On the other hand, it is reflexive since (x, x) ∈ R for all cases of x: x = a, x = b, and x = c. Likewise, it is symmetric since (а, b) ∈ R and (b, а) ∈ R and (b, с) ∈ R and (c, b) ∈ R.

2) Let S=Z and define R = {(x,y) |x and y have the same parity}

i.e., x and y are either both even or both odd.

The parity relation is an equivalence relation.

a. For any x ∈ Z, x has the same parity as itself, so (x,x) ∈ R.

b. If (x,y) ∈ R, x and y have the same parity, so (y,x) ∈ R.

c. If (x.y) ∈ R, and (y,z) ∈ R, then x and z have the same parity as y, so they have the same parity as each other (if y is odd, both x and z are odd; if y is even, both x and z are even), thus (x,z)∈ R.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial but not transitive, so the relation provided in (1) satisfies this condition.

Step-by-step explanation:

1) By definition,

a) R, a relation in a set X, is reflexive if and only if ∀x∈X, xRx ---> xRx.

That is, x works at the same place of x.

b) R is symmetric if and only if ∀x,y ∈ X, xRy ---> yRx

That is if x works at the same place y, then y works at the same place for x.

c) R is transitive if and only if ∀x,y,z ∈ X, xRy∧yRz ---> xRz

That is, if x works at the same place for y and y works at the same place for z, then x works at the same place for z.

2) An equivalence relation on a set S, is a relation on S which is reflexive, symmetric and transitive.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial and not transitive.

QED!

6 0
3 years ago
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