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elena-s [515]
3 years ago
9

Write an equation that represents the line. Use exact numbers

Mathematics
1 answer:
dsp733 years ago
8 0

Answer:

y = 5/6x + 8/3

Step-by-step explanation:

y = mx + b (m is the slope, b is the y-intercept)

(-2, 1) (4, 6)

m = change in y / change in x

m = 5/6

y = 5/6x + b

Plugging in (4, 6):

6 = 5/6(4) + b

6 = 20/6 + b

36/6 = 20/6 + b

36/6 - 20/6 = b

b = 16/6

b = 8/3

y = 5/6x + 8/3

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Which statement describes whether the function is continuous at x = –2?
slava [35]

Answer: D. The function is not continuous at x = –2 because Limit as x approaches negative 2 f(x) does not exist.

Step-by-step explanation:

edge2021

4 0
3 years ago
Read 2 more answers
What is<br> The equation of the line that passes through the points (5,-1) and (4,-5)
Murljashka [212]

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:

y = mx + b

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis

We have two points through which the line passes:

(x_ {1}, y_ {1}) :( 5, -1)\\(x_ {2}, y_ {2}) :( 4, -5)

We found the slope:

m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {-5 - (- 1)} {4-5} = \frac {-5+ 1} {- 1} = \frac {-4} {- 1} = 4

Thus, the equation is of the form:

y = 4x + b

We substitute one of the points and find b:

(x,y):(5,-1)\\-1=4(5)+b\\-1=20+b\\-1-20=b\\b=-21

Finally, the equation is:

y = 4x-21

Answer:

y = 4x-21

4 0
3 years ago
Help?????????????? I need to know the answer
Natasha2012 [34]
The slope of the line is -2
8 0
3 years ago
Use the upper and lower sums to approximate the area of the region using the given number of subintervals (of equal width)
Mashutka [201]
From the figure shown, the interval is divided into 5 equal parts making each subinterval to be 0.2.

Part A:

y= \sqrt{1-x^2}

The approximate the area of the region shown in the figure using the lower sums is given by:

 Area= [y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2]+[y(0.8)\times0.2] \\ +[y(1)\times0.2] \\  \\ =[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2]+[\sqrt{1-(0.6)^2}\times0.2] \\ +[\sqrt{1-(0.8)^2}\times0.2]+[\sqrt{1-(1)^2}\times0.2] \\  \\ =(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2)+(0\times0.2) \\  \\ =0.196+0.183+0.16+0.12=0.659



Part B:

The approximate the area of the region shown in the figure using the lower sums is given by:

 Area= [y(0)\times0.2]+[y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2] \\ +[y(0.8)\times0.2] \\ \\ =[\sqrt{1-(0)^2}\times0.2]+[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2] \\ +[\sqrt{1-(0.6)^2}\times0.2] +[\sqrt{1-(0.8)^2}\times0.2] \\ \\ =(1\times0.2)+(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2) \\ \\ =0.2+0.196+0.183+0.16+0.12=0.859



Part C:

The approximate area of the given region is given by

Area= \frac{0.659+0.859}{2} = \frac{1.518}{2} =0.759
7 0
4 years ago
What is 1 + 4 my good sir
tia_tia [17]

Answer:

5

Step-by-step explanation:

5

lol hope helps :D

6 0
3 years ago
Read 2 more answers
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