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Alexxandr [17]
3 years ago
6

On a particular day, the wind added 3 miles per hour to Alfonso's rate when he was cycling with the the wind and subtracted 3 mi

les per hour from his rate on his return trip. Alfonso found that in the same amount of time he could cycle 48 miles with the wind, he could only go 30 miles against the wind. What is his normal bicycling speed with no wind ?
Mathematics
1 answer:
Mnenie [13.5K]3 years ago
3 0
13 mph

using d=rt you can set up 2 equations

48=(r+3)t
30= (r-3)t
solving these both for t and setting them equal warrants

48/(r+3) = 30/(r-3)
and with cross multiplication and distribution you get 

48r-144=30r+90

solving for r you get r=13
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RideAnS [48]

Question 1:

For this case we have the following equation:

 5x ^ 2 = 6x + 11

We observe that the equation is a polynomial of the second degree.

Therefore, the equation has two possible solutions.

The variable of the equation for this case is x.

In this case, we have an open equation because we have a variable.

Rewriting the equation we have:

5x ^ 2 - 6x - 11 = 0

Answer:

open, there is a variable.


Question 2:

For this case we have the following equation:

 2-8x = -6

From here, we must clear the value of x.

For this, we follow the following steps:

1) Pass the value of 8x adding:

2 = -6 + 8x

2) Pass the value of 6 adding:

2 + 6 = 8x

3) Add the values ​​on the same side of equality:

 8x = 8

4) Pass the 8 to divide:

x = \frac{8}{8} = 1

Answer:

The value of x is given by:

x = 1


Question 3:

For this case we have the following equation:

 y = x-4

The ordered pair solution is the one that satisfies both sides of the equation.

For the point (6, 2) we have:

x = 6  y = 2

Substituting values ​​we have:

2 = 6-4  2 = 2

We observe that both sides of the equation are equal.

Therefore, the ordered pair solution is (6, 2)

Answer:

An ordered pair that is a solution of the equation y = x-4 is:

(6, 2)

3 0
3 years ago
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Is 7849 a reasonable answer for 49 times 49 why or why not
alekssr [168]
No, because when you multiply 49 by itself,you get 2401 not 7849
3 0
3 years ago
Sari is factoring the polynomial 2x2+5x+3. If one factor is (x+1), what is the other factor?
ss7ja [257]

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Step-by-step explanation:

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3 years ago
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Find the average rate of change of the function between the given values of x. y = 9+ 5x + 0.5x2 between x = 2 and x = 4.
svet-max [94.6K]

Answer:

The average rate of change is 8.

Step-by-step explanation:

The formula to calculate the average rate of change of a function F(x) is:

\frac{F(b)-F(a)}{b-a}

In this case, F(x) = 0.5x^{2} +5x+9

a=2 and b=4

You have to evaluate x=2 (which is a in the formula) and x=4 (which is b in the formula) in the function.

In order to obtain F(b) and F(a) you have to replace x=4 and x=2 in the given function:

F(b) = (0.5)4^{2} + 5(4) +9= 37

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3 0
3 years ago
1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff
Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If  A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that  f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0
3 years ago
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