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babunello [35]
2 years ago
9

How do you solve this?

Mathematics
2 answers:
Elina [12.6K]2 years ago
8 0

Answer: C: 1/2

Step-by-step explanation:

Rise over run tells us that this graph is going up 1 and right 2. Since it’s going up and right it’s going to be positive and since we do rise over run on our slope fraction, that leaves us with C, 1/2.

NNADVOKAT [17]2 years ago
4 0

Answer:

C.)

Step-by-step explanation:

The graph is positive this time, so right away you can eliminate A and B.

If y=2x, the graph would be much steeper so the answer is C.) you can also check by graphing it in desmos.

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Which function is graphed below?
GalinKa [24]

Answer:

f(x) = sin(x)

3 0
1 year ago
Answer quick please!
8_murik_8 [283]

Answer:

y = x/2 + 5

Step-by-step explanation:

y = x/2 + 5

⇒ y = (1/2)x + 5

which satisfies the linear equation y = mx + b

where m is the slope and b is the y-intercept

8 0
1 year ago
Read 2 more answers
Using the bijection rule to count binary strings with even parity.
AleksandrR [38]

Answer:

Lets denote c the concatenation of strings. For a binary string <em>a</em> in B9, we define the element f(a) in E10 this way:

  • f(a) = a c {1} if a has an odd number of 1's
  • f(a) = a c {0} if a has an even number of 1's

Step-by-step explanation:

To show that the function f defined above is a bijective function, we need to prove that f is well defined, injective and surjective.

f   is well defined:

To see this, we need to show that f sends elements fromo b9 to elements of E10. first note that f(a) has 1 more binary integer than a, thus, it has 10. if a has an even number of 1's, then f(a) also has an even number because a 0 was added. On the other hand, if a has an odd number of 1's, then f(a) has one more 1, as a consecuence it will have an even number of 1's. This shows that, independently of the case, f(a) is an element of E10. Thus, f is well defined.

f is injective (or one on one):

If a and b are 2 different binary strings, then f(a) and f(b) will also be different because the first 9 elements of f(a) form a and the first elements of f(b) form b, thus f(a) is different from f(b). This proves that f in injective.

f is surjective:

Let y be an element of E10, Let x be the first 9 elements of y, then f(x) = y:

  • If x has an even number of 1's, then the last digit of y has to be 0, and f(x) = x c {0} = y
  • If x has an odd number of 1's, then the last digit of y has to be a 1, otherwise it wont be an element of E10, and f(x) = x c {1} = y

This shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.

3 0
3 years ago
jacob bought company stock when it was $62.51 a share. In one year the stock increased to $79.56 a share. how much did the stock
Ket [755]
$79.56- $62.51= $17.05.

The stock increased $17.05 in one year~
6 0
3 years ago
Read 2 more answers
If the profits in your consulting business increase by 5​% one year and decrease by 3​% the following​ year, your profits are up
Leona [35]
Let the original profits be $x. 5% increase in profits will give us:
105/100×x=$1.05x
3% reduction in new profits ill give us the latest amount to be:
97/100×1.05x
=$1.0185x
comparing our new profits to our original amount of $x we shall have new percentage increase to be:
(1.0185x-x)/x×100
=1.85% 
Thus our final increase  1.85%

6 0
2 years ago
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