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Scilla [17]
3 years ago
11

What is the measure of the missing angle? PICTURE BELOW \/

Mathematics
1 answer:
jeka57 [31]3 years ago
6 0

Answer:

The measure of the missing angle is 90°

Step-by-step explanation:

∵ This figure is a quadrilateral

∴ the sum of measures of its angles = 360°

∴ m∠x = 360 - 107 - 72 - 91 = 90°

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THERE ARE 3 BAGS OF APPLES WEIGHING A TOTAL OF 22 1/2 POUNDS. TWO OF THE BAGS WEIGH 6 3/8 POUNDS AND 3 1/4 POUNDS. HOW MUCH DOES
Mkey [24]
I wasn't going to click on this one, but the all-caps enthralled me and hypnotized me.

first off, let's change all the mixed fractions to "improper" and proceed from there, keep in mind that if we subtract the two bags' weight from the total, what's leftover is the 3rd bag's weight.

\bf \stackrel{mixed}{22\frac{1}{2}}\implies \cfrac{22\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{45}{2}}
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\stackrel{mixed}{6\frac{3}{8}}\implies \cfrac{6\cdot 8+3}{8}\implies \stackrel{improper}{\cfrac{51}{8}}
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\stackrel{mixed}{3\frac{1}{4}}\implies \cfrac{3\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{13}{4}}\\\\
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\bf \stackrel{\textit{sum of the two bags}}{\cfrac{51}{8}+\cfrac{13}{4}}\impliedby \textit{our LCD is 8}\implies \cfrac{(1)51+(2)13}{8}
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\cfrac{51+26}{8}\implies \cfrac{77}{8}\\\\
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\stackrel{total}{\cfrac{45}{2}}~-~\stackrel{two~bags}{\cfrac{77}{8}}\impliedby \textit{our LCD is again 8}\implies \cfrac{(4)45-(1)77}{8}
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\cfrac{180~~-~~77}{8}\implies \cfrac{103}{8}\implies \stackrel{third~bag}{12\frac{7}{8}}
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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).

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