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3 years ago

How is completing the square beneficial?

1 answer:

8 0

If it is, then you can solve the equation by taking the square root of both sides of the equation. ... Next, if the coefficient of the squared term is 1 and the coefficient of the linear (middle) term is even, completing the square is a good method to use. Finally, the quadratic formula will work on any quadratic equation

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Graph this parametric equation x = 2t y = t + 5, -2 ≤ t ≤ 3

UNO [17]

We are given the following data: x = 2t, y = t + 5, -2 ≤ t ≤ 3. The data is valid since there are three unknowns in this problem and that three equations would suffice to answer the problem.

We start with the given -2 ≤ t ≤ 3 then substitute y = t +5 by using the limits of the range:

at t = -2 ; y = -2 + 5 = 3
at t = 3, y = 3+5 = 8

for the second equation

at t = -2 ; x = 2*-2 =-4
at t=3; x = 2*3 = 6

we group the points based on their original corresponding t's

(3,-4) and (8,6) we just have to connect these points along with the internal points in between. The relationship should be linear.

6 0

3 years ago

Your car's back window is in the shape of a trapezoid with the dimensions shown. The 1616-inch window wiper cleans a part of the

ExtremeBDS [4]

Semicircle area = circle area / 2 = (pi*r^2)/2 16 in is the radius. Let's call the area of this semicircle s. 3.14*16^2=s

Trapezoid area = ((base 1 + base 2) / 2) * h Here, base 1 is 44 and base 2 is 36. The height (h) is 20. Let's call this trapezoid's area t. (44+36)/2 *20 = t

We can remove the area of the semicircle (that created by the windshield wiper) by subtracting it from the area of the trapezoid. t - s = uncleaned area [(44+36)/2 * 20] - [3.14 * 16^2] = uncleaned area

5 0

2 years ago

Read 2 more answers

What is the value of x in equation (-35)^x = 1

FinnZ [79.3K]

Answer:

x = 0.

Step-by-step explanation:

(-35)^x = 1

Now any number to the power 0 is 1.

So x = 0.

5 0

3 years ago

The model shown represents which of the following?

-BARSIC- [3]

Answer:

its 3

Step-by-step explanation:

selcet 3 it goes plus 3 then back to neg 4 its the 3rd option

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