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inna [77]
3 years ago
9

I dont know the answer to this question do you 3 x blank = 1

Mathematics
2 answers:
faltersainse [42]3 years ago
3 0

Answer:

y = \frac13

Step-by-step explanation:

let the blank be y

3 * y = 1

3y = 1

Now, divide both sides by 3:

3y ÷ 3 = 1 ÷ 3

y = 1/3

-Chetan K

Anon25 [30]3 years ago
3 0

Answer:

<u>Value of the blank </u>(We assumed it as x while solving)<u> :-</u>

\boxed {\tt x =  \cfrac{1}{3}}

<u>Actual answer :-</u>

\boxed{\tt \: 3 \times  \boxed{\cfrac{1}{3}}  = 1}

Step-by-step explanation:

<u>Given equation :-</u>

\sf 3 \times ? = 1

(<u>Note</u>: I wrote "?" instead of "blank".)

<em>We need to find the value of the blank on this equation</em>.

<u>Solution:-</u>

To solve Further let's us assume that the value of the blank be x.

So, blank = x

The equation will be . . .

\sf \longmapsto3 \times x = 1

Let's solve out this equation !

\sf \implies3 \times x = 1

<u>Multiply 3 and x on the LHS :-</u>

\sf \implies3 x = 1

<em>Hence the new equation would be </em>3x=1.

<u>Now, Divide both sides by 3 :-</u>

\sf \implies \:  \cfrac{3x}{3}  =  \cfrac{1}{3}

<u>Simplify this equation :-</u>

Cancel 3 and 3 on the LHS, Leave x . 1/3 can't be cancelled.

\sf \implies \:  \cfrac{ \cancel3}{  \cancel3x}  = \cfrac{1}{3}

\sf \implies1x =  \cfrac{1}{3}

As we know that x = 1x So,

\sf \implies \: x =  \cfrac{1}{3}

<em>This equation can't be simplified more.</em>

Hence, the value of x would be 1/3.

<u>Now,Put the value of </u><u>x</u><u> </u><u>(1/3) on the given equation:-</u>

\sf \implies \:  3 \times  \: \cfrac{1}{3}  \:   = 1

We are done !

______________________________________

I hope this helps !

Let me know if you have any questions.

\Huge\sf :)

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A. Find the amplitude.
Feliz [49]

Answers:

  • a) Amplitude = 2
  • b) Period = pi
  • c) Vertical shift = -2, which means it has been shifted down 2 units.
  • d) Horizontal shift = 3pi/8, this shifting is to the right.
  • e) There is <u>  one  </u> cycle between 0 and 2pi.
  • f) The equation of the graph is y = 2*sin(2(x-3pi/8))-2

========================================================

Explanations:

Part (a)

The highest point is when y = 0 and the lowest point is when y = -4. The vertical distance between the peak and valley is 4 units, which cuts in half to 2. This is the amplitude. It's the vertical distance from the center to either the peak or valley.

Note: Amplitude is always positive as it measures a distance.

---------------------

Part (b)

For x > 0, the first valley or lowest point occurs between 0 and pi/4. It appears to be the midpoint of the two values. So that would be (0+pi/4)/2 = pi/8.

The next valley occurs between pi and 5pi/4. Compute the midpoint to get (pi+5pi/4)/2 = (4pi/4+5pi/4)/2 = (9pi/4)/2 = 9pi/8

So we have the graph go from one valley x = pi/8 to the next valley over x = 9pi/8. This is a distance of pi units because 9pi/8-pi/8 = 8pi/8 = pi

The graph repeats itself every pi units, so the period is pi.

---------------------

Part (c)

The midline is normally through y = 0, aka the x axis. However, the graph shows the midline is through y = -2. This means the graph has been shifted down 2 units.

---------------------

Part (d)

This will depend on whether you use sine or cosine. This is entirely because cosine is a phase-shifted version of sine, and vice versa. I'll go with sine.

The parent sine function y = sin(x) goes through the origin (0,0)

However, as part (c) mentioned, we shifted the graph 2 units down. So we have y = sin(x)-2. But plugging x = 0 into this leads to the point (0,-2)

This doesn't match what the graph says. The graph shows the point (3pi/8, -2) on the red curve. The x coordinate 3pi/8 is the midpoint of pi/4 and pi/2

This must mean we need to shift the sine graph 3pi/8 units to the right.

---------------------

Part (e)

Start at the lowest point when x = pi/8. If you start the cycle here, then it ends when x = 9pi/8. See part (b).

So far we've completed one cycle. If we start at x = 9pi/8, then the next valley or lowest point is slightly beyond or to the right of x = 2pi. This means we run out of room and we haven't completed a full cycle.

Overall, one full cycle is between 0 and 2pi.

---------------------

Part (f)

Again I'm going to use sine instead of cosine. Refer back to part (d).

The general sine curve equation is

y = A*sin(B(x-C))+D

where

  • |A| = amplitude
  • B handles the period, specifically T = 2pi/B where T is the period. We can solve for B to get B = 2pi/T
  • C = horizontal phase shift
  • D = vertical shift, and ties together with the midline equation

In this case, we found that

  • |A| = 2
  • T = pi leads to B = 2pi/T = 2pi/pi = 2
  • C = 3pi/8
  • D = -2

So,

y = A*sin(B(x-C))+D

will update to

y = 2*sin(2(x-3pi/8))-2

which is one way to express the equation of the red curve. Optionally you can distribute the 2 through to (x-3pi/8).

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