All categories

4 years ago

SHOW YOUR WORK!!

Mathematics

1 answer:

Dmitrij [34]4 years ago

5 0

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

You might be interested in

X=-9+6 helpppp!!!!!!!

shutvik [7]

Answer:

-3

Step-by-step explanation:

First, we have to do -9 + 6.

The order of addition does not matter, so we can switch around the numbers, making the statement:

6+ (-9)

If this is the equation, we can cancel out the addition sign:

6-9

Now we have to subtract 9 from six, and we will get -3.

x = -3

6 0

3 years ago

Read 2 more answers

7th and 8th graders at Marshall Middle are going on a field trip. The 7th graders use 10 small buses and 12 large buses for 504

Hunter-Best [27]

Answer:

by dividing 325 by 48.

The answer comes out to 6.77

But since the number of buses must be a whole number, we will need 7 buses to carry the 325 students and teachers on a field trip.

so the answer is 7 buses

3 0

3 years ago

A juice manufacturing company designed a carton as shown below. What is the name of the three-dimensional

kodGreya [7K]

Answer: rectangular prism and a pyramid I think

Step-by-step explanation:

7 0

3 years ago

The coordinate plane below represents a city.

alex41 [277]

Answer:

Part A) The system of inequalities is

$x\geq2$ and $y\geq2$

Part B) In the procedure

Part C) The schools that Natalie is allowed to attend are A,B and D

Step-by-step explanation:

Part A: Using the graph above, create a system of inequalities that only contain points C and F in the overlapping shaded regions

we have

Points C(2,2), F(3,4)

The system of inequalities could be

$x\geq2$ -----> inequality A

The solution of the inequality A is the shaded area at the right of the solid line x=2

$y\geq2$ -----> inequality B

The solution of the inequality B is the shaded area above of the solid line y=2

see the attached figure N 1

Part B: Explain how to verify that the points C and F are solutions to the system of inequalities created in Part A

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities

Verify point C

C(2,2)

Inequality A

$x\geq2$ -----> $2\geq2$ ----> is true

Inequality B

$y\geq2$ ------> $2\geq2$ ----> is true

therefore

Point C is a solution of the system of inequalities

Verify point D

F(3,4)

Inequality A

$x\geq2$ -----> $3\geq2$ ----> is true

Inequality B

$y\geq2$ ------> $4\geq2$ ----> is true

therefore

Point D is a solution of the system of inequalities

Part C: Natalie can only attend a school in her designated zone. Natalie's zone is defined by y < −2x + 2. Explain how you can identify the schools that Natalie is allowed to attend.

we have

$y < -2x+2$

The solution of the inequality is the shaded area below the dotted line $y=-2x+2$

The y-intercept of the dotted line is the point (0,2)

The x-intercept of the dotted line is the point (1,0)

To graph the inequality, plot the intercepts and shade the area below the dotted line

see the attached figure N 2

therefore

The schools that Natalie is allowed to attend are A,B and D

6 0

4 years ago

How long is a guy wire reaching from the top of a 21ft. Pole to a point on the ground 15ft. From the pole

vovikov84 [41]

Answer:

26ft

Step-by-step explanation:

This would form a triangle so you would use Pythagorean theorem to find the hypotenuse ( the longest side )

5 0

3 years ago