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garri49 [273]
3 years ago
13

Which table represents a linear function?

Mathematics
2 answers:
motikmotik3 years ago
7 0

Answer:

It’s D

Step-by-step explanation:

Paladinen [302]3 years ago
4 0

Answer:

2.

Step-by-step explanation:

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For each equation, decide if it is always true or never true. x − 13 = x + 1 x + 1 2 = x − 1 2 2 ( x + 3 ) = 5 x + 6 − 3 x x − 3
mrs_skeptik [129]

"x" (and any subsequent words) was ignored because we limit queries to 32 words.

6 0
3 years ago
Periodic grab some phase shifts
nikitadnepr [17]

Answer:

The sine function is:

y(x)=\frac{2}{3}sin[3\pi (x-\frac{2\pi}{3})]-2

Step-by-step explanation:

We need to recall that a sine function can write as:

y(x)=Asin[B(x+C)]+D

Where:

  • A is the amplitude of the function
  • B is the period
  • C is the phase shift left
  • D is the vertical shift

So we just need to use the listed attributes

Therefore, the sine function is:

y(x)=\frac{2}{3}sin[3\pi (x-\frac{2\pi}{3})]-2

I hope it helps you!

8 0
3 years ago
No link or bot answer the question
sesenic [268]

Hello there! :)

\sqrt{81}

It's rational

it's an integer

Hope it helps!

~Just an emotional teen listening to her favorite song "Try Everything"

SilentNature

6 0
3 years ago
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patrick is thinking of a number. his number is twice as large as the smallest number that rounds to 30 when rounding to the near
kati45 [8]
I think the answer is 50 because the smallest number that rounds to 30 is 25 and twice 25 is 50
8 0
3 years ago
Find side lengths C and H
Nuetrik [128]

Answer:

\displaystyle c=\frac{2\sqrt{3}}{3}\\\\\displaystyle h=7\sqrt{2}\\

Step-by-step explanation:

<u>Right Triangles</u>

A right triangle can be identified by the fact it has an internal angle of 90°. In a right triangle, the trigonometric ratios stand.

Let's consider the triangle to the left.  We need to calculate side c, one of the legs of the triangle. We can use the angle adjacent to it (60°) or the angle opposite to it (30°) with the appropriate trigonometric ratio.

We'll use the adjacent angle, and

\displaystyle tan60^o=\frac{2}{c}

Solving for c

\displaystyle c=\frac{2}{tan60^o}=\frac{2}{\sqrt{3}}

Rationalizing

\displaystyle c=\frac{2}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}

Now for the triangle to the right. The side h is the hypotenuse. Again, any of the two angles can be used (though they are equal, for it's an isosceles triangle). For any of them it is true that

\displaystyle sin45^o=\frac{7}{h}

Solving for h

\displaystyle h=\frac{7}{sin45^o}=\frac{7}{\frac{\sqrt{2}}{2}}=\frac{14}{\sqrt{2}}

Rationalizing

\displaystyle h=\frac{14}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{14\sqrt{2}}{2}=7\sqrt{2}

3 0
4 years ago
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