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3 years ago

How to take a line in desmos and reflect it over y axis and find equation.

Mathematics

1 answer:

Pachacha [2.7K]3 years ago

7 0

Answer:

Reflecting a function over any axis can be complicated if you do not know the proper way to do it, so I am going to use examples to show you.

If f(x)=x is your normal function, then flipping it across the y-axis would look like f(x)=-x.

Step-by-step explanation:

Placing a - sign before the X will make it flip across the y-axis. After the X, and it flips across the X-axis.

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10 points!! plzzzzzzz helpppp :( find the letters ( K,K,T )

zheka24 [161]

Answer:

dude u lie it says 5 points

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Will give brainliest.

In-s [12.5K]

Answer:

y = 3x -8

Step-by-step explanation:

I find it convenient to start with a version of the point-slope form of the equation for a line. That is, for point (h, k) and slope m, ...

y = m(x -h) +k

For your m=3 and (h, k) = (3, 1), this equation becomes ...

y = 3(x -3) +1

Eliminating parentheses puts this in the form you desire:

y = 3x -8

6 0

3 years ago

Help me please thanks

kow [346]

Correct answer:
First reflect across the y-axis, then rotate 90 degrees clockwise about point K, then shift 3 units down.

6 0

3 years ago

CALCULUS: For an object whose velocity in ft/sec is given by v(t) = sin(t), what is its distance, in feet, travelled on the inte

rodikova [14]

The linked answer is wrong because that integral gives you the net displacement of the object, not the total distance.

To get the distance, you have to integrate the speed (as opposed to velocity), which involves integrating the absolute value of the velocity function.

$\mathrm{distance} = \displaystyle\int_1^5 |\sin(t)| \,\mathrm dt$

By definition of absolute value,

$|\sin(t)|=\begin{cases}\sin(t)&\text{for }\sin(t)\ge0\\-\sin(t)&\text{for }\sin(t)$

Over this particular integration interval,

• sin(t ) ≥ 0 for 1 ≤ t < π, and

• sin(t ) < 0 for π < t ≤ 5

so you end up splitting the integral at t = π as

$\mathrm{distance} = \displaystyle\int_1^\pi \sin(t)\,\mathrm dt + \int_\pi^5 (-\sin(t))\,\mathrm dt$

Now compute the distance:

$\mathrm{distance} = -\cos(t)\bigg|_1^\pi + \cos(t)\bigg|_\pi^5$

$\mathrm{distance} = -(\cos(\pi) - \cos(1)) + (\cos(5) - \cos(\pi))$

$\mathrm{distance} = -2\cos(\pi) + \cos(1) + \cos(5) \approx 2.82$

making B the correct answer.

7 0

2 years ago

A jewelry shop sells 240 necklaces in a month.

zhuklara [117]

Answer:3 to 2

Step-by-step explanation:

5 0

2 years ago