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3 years ago

Does this graph represent a function? Why or why not?

Mathematics

1 answer:

Yuliya22 [10]3 years ago

5 0

Answer:

It’s D because no lines overlap so it passes the vertical line test

Step-by-step explanation:

You might be interested in

Solve for x. 3/8(2x+16)−2=13

Goryan [66]

3/8(2x+16)-2 = 13
3(2x+16)/8-2 = 13
6x+48/8-2 = 13
6x+48-16/8 = 13
6x+48-16 = 13×8
6x+48-16 = 104
6x = 104+16-48
6x = 120-48
6x = 72
x = 72/6
x = 12

3 0

4 years ago

5 times a number increased by 18 is the same as 36 more than times 4 times the number. Find the number

Tju [1.3M]

Answer:

Step-by-step explanation:

Let's assume that that question is meant to read:

________________________________________________

" 5 times a number increased by 18 is the same as 36 more than 4 times the number. Find the number ".

_______________________________________________

Let the variable "x" (i.e. the lower-case letter ex]—represent the unknown number—for which we are asked to solve.

If the question/problem is meant to read:

_________________________________

" 5 times a number increased by 18 is the same as 36 more than 4 times the number. Find the number ".

_________________________________

Then, treat the problem as:

_________________________________

" 5 times [a number increased by 18] " ; {is the same as}: " 36 more than [4 times the number]". Find the number ".

_________________________________

Note that: " {is the same as:}" ; means: "equals:} ;

So: "...[a number; that is, an unknown number; that is, "x" ; increased by "18" ]" ; would be represented by: "[x + 18]" .

5 times that value would be represented as:

→ 5* (x + 18) ; or: " 5(x + 18) " .

Then we add the: " = " ["equals"] sign:

_________________________________

Then, we consider: "... 36 more than [4 times the number]" .

4 times the [unknown number]: would be written as:

→ " 4 * x " ; or simply: " 4x " .

→ "36 more than this [the above value; i.e. "4x" ; would be represented by adding "36" to said value; as follows:

→ " 4x + 36 " ;

_________________________________________

Now we can:

1) Write our expression as an equation; and then:

2) Solve for the value for "x" ; our unknown number.

___________________________________________

Here is the expression, as an equation:

________________________________________

→ 5(x + 18) = 4x + 36 ;

Now, solve for "x" ;

_________________________________________

Start with the "left-hand side" of the equation:

→ 5(x + 18) ;

Let us expand this expression.

___________________________________

Note the "distributive property" of multiplication:

→ a(b+c) = ab + ac .

___________________________________

As such: " 5(x + 18) = (5 * x) + (5 * 18) " = 5x +120 .

Now, rewrite the equation:

→ 5x + 120 = 4x + 36 ;

Let us subtract "36" from each side of the equation; & subtract "4x" from each side of the equation:

→ 5x + 120 = 4x + 36 ;

- 4x - 36 = - 4x - 36

________________________

to get: 1x + 54 = 10 ;

________________________________________

Rewrite as: " x + 54 = 10 " ;

→ {Since: "1x = x " ;

→ {since: (" 1 * [any numerical value] = that same numerical value"};

Note that this refers to the "identity property of multiplication."

___________________________________________

→

_________________________________________

→ We have: " x + 54 = 10 " ; Solve for "x" ;

Subtract "54" from Each Side of the equation;

to isolate "x" on one side of the equation; & to solve for "x" ;

→ " x + 54 - 54 = 10 - 54 " ;

to get: " x = -44 " .

_______________________________

Answer: The number is " - 44 " .

_______________________________

Let us substitute this value into our equation; to check our work:

→ 5(x + 18) = 4x + 36 ;

→ 5(-44 + 18) ≟ 4(-44) + 35 ?? ;

→ 5(-26) ≟ -176 + 35 ?? ;

→ -130 ≟ -176 + 35

3 0

3 years ago

Write a sine and cosine function that models the data in the table. I need steps to both the sine and cosine functions for a, b,

dangina [55]

Answer(s):

$\displaystyle y = -29sin\:(\frac{\pi}{6}x + \frac{\pi}{2}) + 44\frac{1}{2} \\ y = -29cos\:\frac{\pi}{6}x + 44\frac{1}{2}$

Step-by-step explanation:

$\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 44\frac{1}{2} \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-3} \hookrightarrow \frac{-\frac{\pi}{2}}{\frac{\pi}{6}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{12} \hookrightarrow \frac{2}{\frac{\pi}{6}}\pi \\ Amplitude \hookrightarrow 29$

OR

$\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 44\frac{1}{2} \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{12} \hookrightarrow \frac{2}{\frac{\pi}{6}}\pi \\ Amplitude \hookrightarrow 29$

You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of sine, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the centre photograph displays the trigonometric graph of $\displaystyle y = -29sin\:\frac{\pi}{6}x + 44\frac{1}{2},$ in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the cosine graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the sine graph [centre photograph] is shifted $\displaystyle 3\:units$ to the right, which means that in order to match the cosine graph [photograph on the left], we need to shift the graph BACKWARD $\displaystyle 3\:units,$ which means the C-term will be negative, and by perfourming your calculations, you will arrive at $\displaystyle \boxed{3} = \frac{-\frac{\pi}{2}}{\frac{\pi}{6}}.$ So, the sine graph of the cosine graph, accourding to the horisontal shift, is $\displaystyle y = -29sin\:(\frac{\pi}{6}x + \frac{\pi}{2}) + 44\frac{1}{2}.$ Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph WILL hit $\displaystyle [12, 15\frac{1}{2}],$ from there to the y-intercept of $\displaystyle [0, 15\frac{1}{2}],$ they are obviously $\displaystyle 12\:units$ apart, telling you that the period of the graph is $\displaystyle 12.$ Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at $\displaystyle y = 44\frac{1}{2},$ in which each crest is extended twenty-nine units beyond the midline, hence, your amplitude. Now, there is one more piese of information you should know -- the cosine graph in the photograph farthest to the right is the OPPOCITE of the cosine graph in the photograph farthest to the left, and the reason for this is because of the negative inserted in front of the amplitude value. Whenever you insert a negative in front of the amplitude value of any trigonometric equation, the whole graph reflects over the midline. Keep this in mind moving forward. Now, with all that being said, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

I am delighted to assist you at any time.

4 0

3 years ago

(a)

yKpoI14uk [10]

Part a) is b part b) is b to

5 0

4 years ago

What are factors of 58

Aleonysh [2.5K]

Bruh google it XD
______________

6 0

3 years ago