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svet-max [94.6K]
3 years ago
10

A line passes through the points (15,−13) and (16,−11). Hollis writes the equation y+13=(x−15) to represent the line. Which answ

er correctly analyzes his equation? His equation is incorrect. He incorrectly wrote the slope and switched the values of the coordinates. He should have written y−15=2(x+13). His equation is incorrect. He switched the values of the coordinates. He should have written y−15=x+13. His equation is correct. He correctly included and placed all parts for the point-slope form of the equation of the line. His equation is incorrect. He switched the signs of the coordinates. He should have written y−13=x+15. His equation is incorrect. He incorrectly wrote the slope and switched the signs of the coordinates. He should have written y−13=2(x+15). His equation is incorrect. He incorrectly wrote the slope in his equation. He should have written y+13=2(x−15).
Mathematics
1 answer:
Andru [333]3 years ago
8 0

The equation formula is y - y1 = m(x-x1)

Using the first point for x1, y1:

Y +13 = m(x-15)

M is the slope which is the change in y over the change in x:

M = -11–13 / 16-15 = 2/1 = 2

The equation becomes y +13 =2(x+15)

The answer is:

He incorrectly wrote the slope in his equation. He should have written y+13=2(x−15).

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The hypotenuse of a right triangle is 50 millimeters long. One leg of the right triangle is 30 millimeters long. What is the len
agasfer [191]
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4 years ago
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The range of which function is (2, infinity)? y = 2x y = 2(5x) y = 5x +2 y = 5x + 2
Sveta_85 [38]

Answer:

I think your functions are y=2^{x} ,y=2*5^{x} and y=2+5^{x}

If yes then then the third function which is y=2+5^{x}.

Step-by-step explanation:

The function c^{x} where c is a constant has

Domain : c\geq 0

Range : ( 0 , ∞ )

The above range is irrespective of the value of c.

I have attached the graph of each of the function, you can look at it for visualization.

  • <em>y=2^{x} ⇒ </em>This function is same as  c^{x} so its range is <em>( 0 , ∞ )</em>.
  • <em>y=2*5^{x} ⇒ </em>If we double each value of the function y=5^{x}, which has range ( 0 , ∞ ), but still the value of extremes won't change as 0*2=0 and ∞*2=∞. Therefore the range remains as <em>( 0 , ∞ )</em>.
  • <em>y=2+5^{x}</em> ⇒ If we add 2 to each value of the function y=5^{x}, which has range ( 0 , ∞ ), the lower limit will change as 0+2=2 but the upper limit will be same as ∞. Therefore the range will become as <em>( 2 , ∞ )</em>.

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Tell me these answers
Oksana_A [137]

Answer:

1/3 = 3/9

1/2 = 10/20

3/4 = 12/16

8 0
3 years ago
-16 + 5n= -7(-6 + 8n) + 3
Amiraneli [1.4K]

\text{Hello there!}\\\\\text{In this question, we're trying to solve the equation}\\\text{We would use the distributive property and solve after that}\\\\\text{Lets solve:}\\\\-16 + 5n= -7(-6 + 8n) + 3\\\\\text{Distribute the -7 to the numbers inside the parenthesis}\\\\-16 + 5n= 42 -56n + 3\\\\\text{Add 56n to both sides}\\\\-16 + 61n= 42  + 3\\\\\text{Add 16 to both sides}\\\\61n= 42  + 3 + 16\\\\61n= 42  + 3 + 16\\\\61n= 45 + 16\\\\61n = 61\\\\\text{Divide both sides by 61}

n = 1

<h3>I hope this helps!</h3><h3>Best regards,</h3><h3>MasterInvestor</h3>
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