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ollegr [7]
3 years ago
13

What is the solution to the linear equation?

Mathematics
1 answer:
GenaCL600 [577]3 years ago
7 0

Answer:

y= 10

Step-by-step explanation:

2.8y+0.2y-5y = -14-16

-2y/-2 = -20/-2

y = 10

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Read 2 more answers
Suppose that the universal set is U={1,2,3,4,5,6,7,8,9,10}. Express each of the following subsets with bit strings (of length 10
Vladimir [108]

Answer:

0011100000

1010010001

0111001110

Step-by-step explanation:

As the question is not complete, Here is the complete question.

Suppose that the universal set is U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Express each of these sets with bit strings where the ith bit in the string is 1 if i is in the set and 0 otherwise.

a) {3, 4, 5}

b) {1, 3, 6, 10}

c) {2, 3, 4, 7, 8, 9}.

So, we need to express a) b) and c) into bit strings.

Firstly, number of elements in the universal set represent the number of bits in the bit string.

Secondly, 1 = yes element is present in both universal set as well as in sub set.

0 = No, element is not present in sub set but present in universal set.

Hence, we have:

a) Sub set {3,4,5} = 0011100000  (As there are 3 1's which means only 3,4,5 are present in both universal set and subset.

Similarly,

b) Sub set {1, 3, 6, 10} = 1010010001

c) Sub set {2, 3, 4, 7, 8, 9} = 0111001110

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To construct a perpendicular for \over{AB}, we must first take a compass & take the distance of its arms wider than half the length of \over{AB}.

This is done in order to get two intersecting arcs in the top & bottom of \over{AB} so that a perpendicular bisector can be drawn through it.

After two intersecting lines are drawn below & above \over{AB}, draw a line joining these 2 points through their points of intersection. The point where it intersects \over{AB} is the middle-most point of \over{AB} & now a perpendicular bisector of \over{AB} is constructed.

\rule{150pt}{2pt}

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62unit/hour is the answer
distance/time
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