All categories

2 years ago

Complete the table for each function. Then

Mathematics

1 answer:

Aleks04 [339]2 years ago

6 0

A=ba ca
E= ce de
I don’t know

You might be interested in

Jacob is ordering custom T-shirts for his soccer team. Long-sleeved shirts cost $15 each and short-sleeved shirts cost $10 each.

Harrizon [31]

Answer:

15x+10y≤250

Step-by-step explanation:

you replace the variables with the number of each shirt and times those numbers together then make sure the statement is true

4 0

4 years ago

Read 2 more answers

If n^3=x and n^4=20x , where n>0, what is the value of x?

Natali5045456 [20]

Step-by-step explanation:

∠2 and ∠7 are alternate exterior angles so they are equal.

So you can create the equation

40x-48 = 20x+52 Subtract 20x from both sides. Add 48 to both sides.

20x = 100

x = 5

if you need m∠2 Substtute 5 for x in 20x+52

20(5) + 52 = 100 +52 m∠2= 152°

m∠7 40(5) - 48 = 200-48 m∠7= 152°

6 0

3 years ago

If K is the midpoint of JL, JK = 8x + 11 and KL = 14x – 1, find JL.

masha68 [24]

Answer:

$JL=54$

Step-by-step explanation:

We are given that K is the midpoint of JL. Using this information, we want to find JL.

By the definition of midpoint, this means that:

$JK=KL$

Substitute them for their equations:

$8x+11=14x-1$

Solve for x. Subtract 8x from both sides:

$11=6x-1$

Add 1 to both sides:

$6x=12$

And divide both sides by 6. Hence:

$x=2$

JL is the sum of JK and KL. Hence:

$JK+KL=JL$

Since JK = KL, substitute either one for the other:

$JK+(JK)=2JK=JL$

Substitute JK for its equation:

$2(8x+11)=JL$

Since we know that x = 2:

$2(8(2)+11)=2(16+11)=2(27)=54=JL$

Thus:

$JL=54$

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,

babymother [125]

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.

5 0

3 years ago

I am confused on numbers 25 and 29, the instructions are at the top.

kvasek [131]

#25 is fairly simple. Plug in -4 and 3 into the equation, and the extraneous root will be the one that does not work.
$\sqrt{12-(-4)} = \sqrt{16} = \frac{+}{}4$
Extraneous root in this case is positive four since +4≠-4
 $\sqrt{12-3} = \sqrt{9} = \frac{+}{}3$
In this case it's negative 3, since -3≠3

#29 can be turned into a quadratic equation.
$x= \sqrt{2x+3}$
Square both sides to get
$x^{2}=2x+3$
Then bring the 2x+3 to the other side, setting the quadratic equal to zero.
$x^{2}-2x-3=0$
Factor to find that it's equivalent to
(x-3)(x+1)=0
Therefore x is equal to positive 3 and negative 1. Plug both back into the original equation. Whichever does not work is the extraneous root, and the answer is the one that does.
 $x= \sqrt{2x+3}$
 $3= \sqrt{2(3)+3}$
 $3= \sqrt{9}$
Extraneous root would be negative 3.

 $-1= \sqrt{2(-1)+3}$
 $-1= \sqrt{1}$
Extraneous root would be positive 1.

Your answers are positive 3 and negative 1.
Extraneous roots are negative 3 and positive 1.

3 0

3 years ago