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tresset_1 [31]
3 years ago
7

10 ft

Mathematics
1 answer:
muminat3 years ago
3 0
10+8+8.9+4=30.9
Rounded to the nearest tenth is 30ft
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The Juarez family is making a cross- country trip on Saturday they traveled 450.8 miles on Sunday they traveled 604.6 miles how
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Step-by-step explanation:$;$$;!’

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Wendy scores 24 points in a basketball game. If all of her baskets were 3-point shots, how many baskets did she make?
yarga [219]
Hello!

If all her shots were 3 pointers we will just divide

24÷3=8

Wendy made 8 baskets.

I hope this helps!
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Which of the following numbers is divisible by 6?
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The answer is a because when u divide that number by 6 u get a whole number
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Solve for area of the shaded region (#61) <br> Pls explain im so confused
BartSMP [9]

Answer: So after reading my explanation you can simply find the area good luck!

Step-by-step explanation:

Inequality 1 is graphed in blue and inequality 2 is graphed in red. The overlap of the shaded regions (purple shading) represents the solution. ... Inequality 1 is graphed in blue and inequality 2 is graphed in red. In this case the regions do not overlap. This indicates that there is no solution to the system

3 0
4 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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