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2 years ago

Why is (3, -5) not a solution to -x+4=-15 and -2x-3y=-8

Mathematics

2 answers:

Sergeeva-Olga [200]2 years ago

8 0

Answer:

see below

Step-by-step explanation:

Check the solution in both equation

-x+4 = -15

-3 +4 = -15

1 does not equal -15 so it is not a solution

-2x-3y = -8

-2(3)+-3(-5) = -8

-6 +15 = -8

9 = -8

It is not a solution to either equation

wlad13 [49]2 years ago

4 0

Step-by-step explanation:

-x+4y= -15

-2x-3y=-8

the intersection point isn't (3,-5)

it's approximately (-1.1818 ,3.45455 )

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Ede4ka [16]

Step-by-step explanation:

(a)

Using the definition given from the problem

$f(A) = \{x^2 \, : \, x \in [0,2]\} = [0,4]\\f(B) = \{x^2 \, : \, x \in [1,4]\} = [1,16]\\f(A) \cap f(B) = [1,4] = f(A \cap B)\\$

Therefore it is true for intersection. Now for union, we have that

$A \cup B = [0,4]\\f(A\cup B ) = [0,16]\\f(A) = [0,4]\\f(B)= [1,16]\\f(A) \cup f(B) = [0,16]$

Therefore, for this case, it would be true that $f(A\cup B) = f(A)\cup f(B)$ .

(b)

1 is not a set.

(c)

To begin with

$A\cap B \subset A,B$

Therefore

$g(A\cap B) \subset g(A) \cap g(B)$

Now, given an element of $g(A) \cap g(B)$ it will belong to both sets, therefore it also belongs to $g(A\cap B)$ , and you would have that

$g(A)\cap g(B) \subset g(A \cap B)$ , therefore $g(A)\cap g(B) = g(A \cap B)$ .

(d)

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$g(A) \cup g(b) \subset g(A\cup B)$

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