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3 years ago

What is the equation for shifting the standard sine curve 2 units horizontally?

1 answer:

3 0

Just add the displacement value to variable. If this value is positive, the graph will move to the right, otherwise it will shift to left.

$f(\O)=sen(\O+x)$

Where: x=displacement value

If you notice any mistake in my English, please let me know, because I am not native.

You might be interested in

What should the IUPAC name for a binary covalent compound lack? prefixes Roman numerals an -ide ending the name of a nonmetal.

vitfil [10]

The IUPAC name for a binary covalent compound lack prefixes such as mono, di, tri tetra etc.

A binary compound is a compound that is composed of only two elements. Many binary compounds could be made up of a metal and a nonmetal or even two nonmetals as the case may be. Examples of binary compounds include; SO2, NaH, K2S etc.

We must note that a binary covalent compound uses Roman numerals to indicate the oxidation state of the central atom in the compound. For instance, SO2 is called Sulfur IV oxide.

Hence, the IUPAC name for a binary covalent compound lack prefixes such as mono, di, tri tetra etc.

Lear more about IUPAC: brainly.com/question/11587934

5 0

2 years ago

Find the difference: -7 -− (-2) = A 9 B -9 C 5 D -5

AveGali [126]

Answer: B. −9

Step-by-step explanation: -7--(-2)

−7+−2

−9

8 0

2 years ago

What is the equation of the line that passes through the points (2, –1) and (6, 1)?

Zolol [24]

Y=kx+b
2k+b=-1
6k+b=1
6k+b-2k-b=2
4k=2
k=1/2
b=-2
y=1/2x-2
So your answer is option A

8 0

2 years ago

.. Which of the following are the coordinates of the vertices of the following square with sides of length a?

atroni [7]

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Step-by-step explanation:

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

To find the sides of a square, let us use the distance formula,

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,a), T(a,a), W(a,0)$ has sides of length a.

Option B: $O(0,0), S(0,a), T(2a,2a), W(a,0)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

This is not a square because the lengths are not equal.

Option C: $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$ has sides of length 2a.

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length OS }=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } S T=\sqrt{(a-a)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } T W=\sqrt{(0-a)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } O W=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

Thus, the square with vertices $O(0,0), S(a,0), T(a,a), W(0,a)$ has sides of length a.

Thus, the correct answers are option a and option d.

8 0

2 years ago

Figure STOP is reflected about the y-axis to obtain figure STO′P: A coordinate plane with two quadrilaterals is shown. Figure ST

gogolik [260]

Answer:

Figure STOP is congruent to figure S′T′O′P′.

Step-by-step explanation:

I just took the test

6 0

3 years ago