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vladimir2022 [97]
3 years ago
6

Which line is linear model for the data

Mathematics
1 answer:
serg [7]3 years ago
7 0
The answer is c......
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Law Incorporation [45]
If it is right triangle use pythagorean
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WHOEVER IS RIGHT GETS BRAINLIST
pochemuha
The answer to the first question is b
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Write an equation that illustrates a number with two decimal places multiplied by a number with one decimal place. The product h
Vilka [71]

Step-by-step explanation:

Let the number with two decimal places be 1.23

Let the number with one decimal place be 4.5

Then 1.23 multiplied by 4.5

= 1.23 × 4.5

= 5.535

The result has more than two nonzero digits.

Note: When you multiply two decimal numbers, the decimal place(s) of the result is the addition of the decimal places of each of the numbers being multiplied.

8 0
3 years ago
Pre cacl pleaseeeee help. the options are in the attachments as well
ivanzaharov [21]

Answer:

The values of x so that y = \csc x have vertical asymptotes are -2\pi, -\pi, 0, \pi, 2\pi.

Step-by-step explanation:

The function cosecant is the reciprocal of the function sine and vertical asymptotes are located at values of x so that function cosecant becomes undefined, that is, when function sine is zero, whose periodicity is \pi. Then, the  vertical asymptotes associated with function cosecant are located in the values of x of the form:

x = 0\pm \pi\cdot i, \forall \,i\in \mathbb{N}_{O}

In other words, the values of x so that y = \csc x have vertical asymptotes are -2\pi, -\pi, 0, \pi, 2\pi.

3 0
3 years ago
Use the parabola tool to graph the quadratic function f(x)=(x-5)^2+1
andrey2020 [161]

Answer:

Look to the attached graph

Step-by-step explanation:

* Lets revise how to graph the quadratic function

- Find the vertex of it

- Find the y-intercept

- Find the x-intercept

∵ f(x) = (x - 5)² + 1 ⇒ the completing square form

- The completing square form for any quadratic is

 ( x - h)² + k, where h and k are the coordinate of the vertex point

* Lets compare the two forms

∵ (x - h)² + k = (x - 5)² + 1

∴ h = 5 and k = 1

∴ The vertex of the parabola is (5 , 1)

- To find the x-intercept put f(x) = 0

∵ (x - 5)² + 1 = 0 ⇒ subtract 1 from both sides

∴ (x - 5)² = -1 ⇒ the square can not give -ve number

∴ The parabola does not intersect the x-axis

- To find the y-intercept put x = 0

∵ f(0) = (0 - 5)² + 1 = 25 + 1 = 26

∴ The parabola intersects the y-axis at point (0 , 26)

- The parabola is opened upward because the coefficient of x² is +ve

* Now lets graph it

- Look to the attached graph

5 0
3 years ago
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