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Orlov [11]
3 years ago
11

10x-5=0 equation ,inverse operation,simplify,inverse operation,solution,check

Mathematics
1 answer:
sattari [20]3 years ago
8 0

Answer:

x=.5

Step-by-step explanation:

First, you need to isolate the variable term, so you have to move -5 over to the other side. Since it is subtraction on this side, you add 5 to both sides to cancel out the -5, and add it to the other side.

Now we have 10x=5. To solve for the variable, we have to divide by the number in the variable term. In this case, it is 10. 10/10 is 1, so now x is alone. We have to do the same thing on the other side. 5/10 is .5.

That's the answer! It's x=.5!

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\Delta KLJ\sim\DeltaNPM (AAA) Corresponding angles are congruent.

Therefore, the sides of the triangles are proportional:

\dfrac{4x-4}{20}=\dfrac{30}{25}    cross multiply

25(4x-4)=(20)(30)    use distributive property

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If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

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7 0
3 years ago
Suppose that X is a subset of Y. Let p be the proposition ‘x is an element ofX’ and let q be the proposition ‘x is an element of
Genrish500 [490]

Answer:

See answer below

Step-by-step explanation:

The statement ‘x is an element of Y \X’ means, by definition of set difference, that "x is and element of Y and x is not an element of X", WIth the propositions given, we can rewrite this as "p∧¬q". Let us prove the identities given using the definitions of intersection, union, difference and complement. We will prove them by showing that the sets in both sides of the equation have the same elements.

i) x∈AnB  if and only (if and only if means that both implications hold) x∈A and x∈B if and only if x∈A and x∉B^c (because B^c is the set of all elements that do not belong to X) if and only if x∈A\B^c. Then, if x∈AnB  then x∈A\B^c, and if x∈A\B^c then x∈AnB. Thus both sets are equal.

ii) (I will abbreviate "if and only if" as "iff")

x∈A∪(B\A) iff x∈A or x∈B\A iff x∈A or x∈B and x∉A iff x∈A or x∈B (this is because if x∈B and x∈A then x∈A, so no elements are lost when we forget about the condition x∉A) iff x∈A∪B.

iii) x∈A\(B U C) iff x∈A and x∉B∪C iff x∈A and x∉B and x∉C (if x∈B or x∈C then x∈B∪C thus we cannot have any of those two options). iff x∈A and x∉B and x∈A and x∉C iff x∈(A\B) and x∈(A\B) iff x∈ (A\B) n (A\C).

iv) x∈A\(B ∩ C) iff x∈A and x∉B∩C iff x∈A and x∉B or x∉C (if x∈B and x∈C then x∈B∩C thus one of these two must be false) iff x∈A and x∉B or x∈A and x∉C iff x∈(A\B) or x∈(A\B) iff x∈ (A\B) ∪ (A\C).

8 0
3 years ago
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