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wlad13 [49]
3 years ago
9

Use the FOIL Method to find (x−4)(x+8).

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
8 0

Answer:

x^2 +4x - 32

Step-by-step explanation:

By FOIL method, (x-4)(x+8) = x \cdot x + x \cdot 8 + -4 \cdot x + -4 \cdot 8 = x^2 +4x - 32.

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they both go up 2 and over 1 (rise over run is slope)
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Use the graph of the function f to find f(-1)
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What are two different ways of factoring –3x – 9? Select all that apply. A. –3(x + 3) B. 3(x + 3) C. 3(–x – 3) D. –3(x – 3)
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8 0
3 years ago
A 20% acid solution is mixed with 20 liters of a 45% acid solution. How much of a 30% acid solution will result?
vaieri [72.5K]

Answer:

50 liters

Step-by-step explanation:

If x is the volume of 20% acid, then:

x (0.20) + 20 (0.45) = (x + 20) (0.30)

0.2x + 9 = 0.3x + 6

3 = 0.1x

x = 30

30 liters of 20% acid are needed, so there will be a total of 50 liters of 30% acid.

6 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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