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Tanya [424]
3 years ago
12

Solve for x. 5 1/4 + x + 6 5/6 + 4 2/3 = 22 1/6 A. 5 5/12 B. 6 1/2 C. 6 7/12 D. 7 1/2

Mathematics
1 answer:
Vladimir79 [104]3 years ago
8 0
<span>5 1/4 + x + 6 5/6 + 4 2/3 = 22 1/6

5 3/12 + x + 6 10/12 + 4 8/12 = 22 2/12
15 21/12 + x = 22 2/13
</span>16 9/12<span> + x = 22 2/12
x = 22 2/12 - </span>16 9/12<span>
x = </span>21 14/12 - 16 9/12
x = 5 5/12

answer is <span>A. 5 5/12</span>
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kvv77 [185]

Answer:

Step-by-step explanation:

The relationship is linear, so the plant grows the same amount each day.

The height on the 2nd day was 8 inches:

h₂ = 8

The height on the 7th day was 20.5 inches:

h₇ = h₂ + (7-2)d = 8 + 5d = 20.5

d = 2.5

The plant grows 2.5 inches each day.

6 0
2 years ago
How would I graph this inequality?<br> <img src="https://tex.z-dn.net/?f=7x%2B3y%5C%20%20%5Ctextless%20%5C%206-5x" id="TexFormul
Anon25 [30]

Answer:

the graph in the attached figure

Step-by-step explanation:

we have

7x+3y

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Subtract 7x both sides

3y

3y

Divide by 3 both sides

y

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The equation of the shaded line is y=-4x+2

The slope of the dashed line is negative m=-4

The y-intercept of the dashed line is the point (0,2)

The x-intercept of the dashed line is the point (0.5,0)

To graph the inequality plot the intercepts and shade the area below the dashed line

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3 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!

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3 years ago
There are 3 cats in a room and no other creatures.each cat has 2 ears, 4 paws, and 1 tail
Nata [24]

So there is a total of 6 ears, 12 paws, and 3 tails in the room.

3 0
2 years ago
Read 2 more answers
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UNO [17]
1-12 is between 5/6 and 13.
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