Find the third term and the common difference. 1, 4, 7, 10

Arnold rides the subway 20 times a month to work for $2.50 a day a monthly pass cost $45.25. How much could he save by buying th

ozzi

Answer:

He would save $4.75.

Step-by-step explanation:

Without the pass,

In a month, Arnold has to spend: $2.50 x 20 = $50

With the pass,

In a month, Arnold has to spend: $45.25

So, Arnold may save: $50 - $45.25 = $4.75 if he has a monthly pass.

7 0

3 years ago

Read 2 more answers

Which of the following are solutions to the system graphed below.

Kisachek [45]

Weird. I think you just need to look if the point falls on the shaded area. But only (-5,5) does ...

4 0

3 years ago

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

3

zlopas [31]

The answer would be C. 5(5) + 2(2) = 29
(5) + 4(2) = 13

3 0

3 years ago

Write three fractions equivalent to \frac{9}{15}

Aleks [24]

(a) On scaling up by a factor of 2 , the equivalent fraction is 18/30.

(b) On scaling up by a factor of 3 , the equivalent fraction is 27/45.

(c) On scaling down by a factor of 3 , the equivalent fraction is 3/5 .

In the question ,

a fraction 9/15 is given ,

Part(a)

to scale up the fraction by a factor of 2 , we multiply the numerator and denominator by 2 .

On multiplying , we get

9/15 = (9*2)/(15*2) = 18/30 .

Part(c)

to scale up the fraction by a factor of 3 , we multiply the numerator and denominator by 3 .

On multiplying , we get

9/15 = (9*3)/(15*3) = 27/45 .

Part(c)

to scale down the fraction by a factor of 3 , we divide the denominator and numerator by 3 .

On dividing , we get

9/15 = (9/3)/(15/3) = 3/5 .

Therefore , (a) On scaling up by a factor of 2 , the equivalent fraction is 18/30.

(b) On scaling up by a factor of 3 , the equivalent fraction is 27/45.

(c) On scaling down by a factor of 3 , the equivalent fraction is 3/5 .

The given question is incomplete , the complete question is

Write three fractions equivalent to 9/15 .

Scale the numerator and denominator up for the first two and down for the third.

(a) Scale up by: __ Equivalent fraction: __

(b) Scale up by: __ Equivalent fraction: __

(c) Scale down by: __ Equivalent fraction: __

Learn more about Equivalent Fraction here

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7 0

1 year ago