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3 years ago

Is the following not a function ?

Mathematics

2 answers:

ExtremeBDS [4]3 years ago

6 0

The answer is Yes it a function

il63 [147K]3 years ago

4 0

Answer:

Yes

Step-by-step explanation:

The "x" is not supposed to be the same. The input has to be different.

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ValentinkaMS [17]

Multiples of 50 are 3000, 450, 50, and 500 and the rest are not multiples of 50

8 0

3 years ago

Read 2 more answers

In the diagram below, de is parallel to xy. what is the value of y

Setler79 [48]

The answer to your problem is D:86

6 0

3 years ago

11. Tickets to a football game were $5 for

slamgirl [31]

Answer:

The number of children that attended the game was 365

Step-by-step explanation:

Let

x ----> the number of children that attended the game

y ---> the number of adults that attended the game

we know that

$x+y=587$ ----> equation A

$5x+7y=3,379$ ----> equation B

Solve the system by graphing

Remember that the solution of the system is the intersection point both graphs

The solution is the point (365,222)

see the attached figure

therefore

The number of children that attended the game was 365

4 0

3 years ago

Question 6 (5 points)

Studentka2010 [4]

Answer:

30 inches

Step-by-step explanation:

using a^2+b^2=c^2

72^2=5184

30^2=900

5184+900=6084

√6084=78

3 0

3 years ago

Looking at the two quadratic functions below (1 & 2), answer the following questions.

algol [13]

Part A:

Given parabola (1) to be $f(x) =-(x+12)^2 -6$ , and parabola (2) to be $f(x)=13(x-4)^2+1$

Notice that parabola (2) is stretched horizontally by a factor of 13 which is greater than 1. This means that parabola (2) is further away from the x-axis than parabola (1). (i.e. parabola (2) is more 'vertical' than parabola (1).

Therefore, parabola (1) is wider than parabola (2).

Part B:

A parabola open up when the coefficient of the quadratic term (the squared term) is positive and opens down when the coefficient of the quadratic term is negative.

Given parabola (1) to be $f(x) =-(x+12)^2 -6$ , and parabola (2) to be $f(x)=13(x-4)^2+1$

Notice that the coefficient of the quadratic term is positive for parabola (2) and negative for parabola (1), therefore, the parabola that will open down is parabola (1).

Part C:

For any function, f(x), the graph of the function is moved p places to the left when p is added to x (i.e. f(x + p)) and moves p places to the right when p is subtracted from x (i.e. f(x - p)).

Given parabola (1) to be $f(x) =-(x+12)^2 -6$ , and parabola (2) to be $f(x)=13(x-4)^2+1$

Notice that in parabola (1), 12 is added to x, which means that the graph of the parent function is shifted 12 places to the left while in parabola (2), 4 is subtracted from x, which means that the graph of the parent function is shifted 4 places to the right.

Therefore, the parabola that would be furthest left on the x-axis is parabola (1).

Part D:

For any function, f(x), the graph of the function is moved q places up when q is added to the function (i.e. f(x) + q) and moves q places down when q is subtracted from the function (i.e. f(x) - q).

Given parabola (1) to be $f(x) =-(x+12)^2 -6$ , and parabola (2) to be $f(x)=13(x-4)^2+1$

Notice that in parabola (2), 1 is added to the function, which means that the graph of the parent function is shifted 1 place up while in parabola (1), 6 is subtracted from the function, which means that the graph of the parent function is shifted 6 places down.

Therefore, the parabola that would be highest on the y-axis is parabola (2).

7 0

4 years ago