I need help please.

Find the distance that the points is given on the graph ?

EleoNora [17]

From the graph, the given points are (2,-1) and (4,0). Use the distance between two points formula:

$\large \boxed{ \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} } }$

Substitute these points in the formula.

$\large{ \sqrt{ {(2 - 4)}^{2} + {( - 1 - 0)}^{2} } } \\ \large{ \sqrt{ {( - 2)}^{2} + {( - 1)}^{2} } } \\ \large{ \sqrt{4 + 1} \longrightarrow \sqrt{5} }$

Hence, the distance is √5

Answer

√5

7 0

3 years ago

PLS HELPPP MEEE ILL GIVE BRAINLIEST

Arlecino [84]

Answer:

y = 2

Step-by-step explanation:

not sure but should be it

4 0

3 years ago

Read 2 more answers

Fie triunghiul ABC cu unghiul A=90* a) Daca AB= 4 cm si unghiul B=60*, aflati unghiul C, AC,BC; b) Daca AC=12cm sin sinB=0,6 , a

sergeinik [125]

Answer:Si eu am nevoie de răspuns si nu gasesc..

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Help ASAP!!!!!!!!!!!! Show your work!!!!!!!!!!!

Mariulka [41]

Answer:

x = -0.846647 or x = -0.177346 or x = 0.841952 or x = 1.58204

Step-by-step explanation:

Solve for x:

5 x^4 - 7 x^3 - 5 x^2 + 5 x + 1 = 0

Eliminate the cubic term by substituting y = x - 7/20:

1 + 5 (y + 7/20) - 5 (y + 7/20)^2 - 7 (y + 7/20)^3 + 5 (y + 7/20)^4 = 0

Expand out terms of the left hand side:

5 y^4 - (347 y^2)/40 - (43 y)/200 + 61197/32000 = 0

Divide both sides by 5:

y^4 - (347 y^2)/200 - (43 y)/1000 + 61197/160000 = 0

Add (sqrt(61197) y^2)/200 + (347 y^2)/200 + (43 y)/1000 to both sides:

y^4 + (sqrt(61197) y^2)/200 + 61197/160000 = (sqrt(61197) y^2)/200 + (347 y^2)/200 + (43 y)/1000

y^4 + (sqrt(61197) y^2)/200 + 61197/160000 = (y^2 + sqrt(61197)/400)^2:

(y^2 + sqrt(61197)/400)^2 = (sqrt(61197) y^2)/200 + (347 y^2)/200 + (43 y)/1000

Add 2 (y^2 + sqrt(61197)/400) λ + λ^2 to both sides:

(y^2 + sqrt(61197)/400)^2 + 2 λ (y^2 + sqrt(61197)/400) + λ^2 = (43 y)/1000 + (sqrt(61197) y^2)/200 + (347 y^2)/200 + 2 λ (y^2 + sqrt(61197)/400) + λ^2

(y^2 + sqrt(61197)/400)^2 + 2 λ (y^2 + sqrt(61197)/400) + λ^2 = (y^2 + sqrt(61197)/400 + λ)^2:

(y^2 + sqrt(61197)/400 + λ)^2 = (43 y)/1000 + (sqrt(61197) y^2)/200 + (347 y^2)/200 + 2 λ (y^2 + sqrt(61197)/400) + λ^2

(43 y)/1000 + (sqrt(61197) y^2)/200 + (347 y^2)/200 + 2 λ (y^2 + sqrt(61197)/400) + λ^2 = (2 λ + 347/200 + sqrt(61197)/200) y^2 + (43 y)/1000 + (sqrt(61197) λ)/200 + λ^2:

(y^2 + sqrt(61197)/400 + λ)^2 = y^2 (2 λ + 347/200 + sqrt(61197)/200) + (43 y)/1000 + (sqrt(61197) λ)/200 + λ^2

Complete the square on the right hand side:

(y^2 + sqrt(61197)/400 + λ)^2 = (y sqrt(2 λ + 347/200 + sqrt(61197)/200) + 43/(2000 sqrt(2 λ + 347/200 + sqrt(61197)/200)))^2 + (4 (2 λ + 347/200 + sqrt(61197)/200) (λ^2 + (sqrt(61197) λ)/200) - 1849/1000000)/(4 (2 λ + 347/200 + sqrt(61197)/200))

To express the right hand side as a square, find a value of λ such that the last term is 0.

This means 4 (2 λ + 347/200 + sqrt(61197)/200) (λ^2 + (sqrt(61197) λ)/200) - 1849/1000000 = (8000000 λ^3 + 60000 sqrt(61197) λ^2 + 6940000 λ^2 + 34700 sqrt(61197) λ + 6119700 λ - 1849)/1000000 = 0.

Thus the root λ = (-3 sqrt(61197) - 347)/1200 + 1/60 (-i sqrt(3) + 1) ((3 i sqrt(622119) - 4673)/2)^(1/3) + (19 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(622119) - 4673)^(1/3)) allows the right hand side to be expressed as a square.

(This value will be substituted later):

(y^2 + sqrt(61197)/400 + λ)^2 = (y sqrt(2 λ + 347/200 + sqrt(61197)/200) + 43/(2000 sqrt(2 λ + 347/200 + sqrt(61197)/200)))^2

Take the square root of both sides:

y^2 + sqrt(61197)/400 + λ = y sqrt(2 λ + 347/200 + sqrt(61197)/200) + 43/(2000 sqrt(2 λ + 347/200 + sqrt(61197)/200)) or y^2 + sqrt(61197)/400 + λ = -y sqrt(2 λ + 347/200 + sqrt(61197)/200) - 43/(2000 sqrt(2 λ + 347/200 + sqrt(61197)/200))

Solve using the quadratic formula:

y = 1/40 (sqrt(2) sqrt(400 λ + 347 + sqrt(61197)) + sqrt(2) sqrt(347 - sqrt(61197) - 400 λ + 172 sqrt(2) 1/sqrt(400 λ + 347 + sqrt(61197)))) or y = 1/40 (sqrt(2) sqrt(400 λ + 347 + sqrt(61197)) - sqrt(2) sqrt(347 - sqrt(61197) - 400 λ + 172 sqrt(2) 1/sqrt(400 λ + 347 + sqrt(61197)))) or y = 1/40 (sqrt(2) sqrt(347 - sqrt(61197) - 400 λ - 172 sqrt(2) 1/sqrt(400 λ + 347 + sqrt(61197))) - sqrt(2) sqrt(400 λ + 347 + sqrt(61197))) or y = 1/40 (-sqrt(2) sqrt(400 λ + 347 + sqrt(61197)) - sqrt(2) sqrt(347 - sqrt(61197) - 400 λ - 172 sqrt(2) 1/sqrt(400 λ + 347 + sqrt(61197)))) where λ = (-3 sqrt(61197) - 347)/1200 + 1/60 (-i sqrt(3) + 1) ((3 i sqrt(622119) - 4673)/2)^(1/3) + (19 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(622119) - 4673)^(1/3))

Substitute λ = (-3 sqrt(61197) - 347)/1200 + 1/60 (-i sqrt(3) + 1) ((3 i sqrt(622119) - 4673)/2)^(1/3) + (19 (i sqrt(3) + 1))/(3 2^(2/3) (3 i sqrt(622119) - 4673)^(1/3)) and approximate:

y = -1.19665 or y = -0.527346 or y = 0.491952 or y = 1.23204

Substitute back for y = x - 7/20:

x - 7/20 = -1.19665 or y = -0.527346 or y = 0.491952 or y = 1.23204

Add 7/20 to both sides:

x = -0.846647 or y = -0.527346 or y = 0.491952 or y = 1.23204

Substitute back for y = x - 7/20:

x = -0.846647 or x - 7/20 = -0.527346 or y = 0.491952 or y = 1.23204

Add 7/20 to both sides:

x = -0.846647 or x = -0.177346 or y = 0.491952 or y = 1.23204

Substitute back for y = x - 7/20:

x = -0.846647 or x = -0.177346 or x - 7/20 = 0.491952 or y = 1.23204

Add 7/20 to both sides:

x = -0.846647 or x = -0.177346 or x = 0.841952 or y = 1.23204

Substitute back for y = x - 7/20:

x = -0.846647 or x = -0.177346 or x = 0.841952 or x - 7/20 = 1.23204

Add 7/20 to both sides:

Answer: x = -0.846647 or x = -0.177346 or x = 0.841952 or x = 1.58204

3 0

3 years ago

Michael read 135 pages in 1 1/2 hours, Select all of the rates that are equivalent to this rate

devlian [24]

The rates that are equivalent to this rate is 90 pages/hr

<h3>Rate of change</h3>

The rate is the number of pages Michael read in 1 hour.

If he read 135 pages in 1 1/2 hours then;

135 pages = 1.5 hours

To determine the rate, we will have:

x = 1 hour

Divide both expressions:

135/x = 1.5/1

1.5x = 135

x = 135/1.5

x = 90

Hence the rates that are equivalent to this rate is 90 pages/hr

Learn more on rate here: brainly.com/question/119866

6 0

2 years ago