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3 years ago

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What is cubic spline?

2 answers:

Mila [183]3 years ago

7 0

Answer:

A cubic spline is a spline constructed of piecewise third-order polynomials in which pass through a set of control points. __________________________________________________________

Hope this helps!!

If I am wrong, please tell me, I enjoy learning from my mistakes:)

lara [203]3 years ago

4 0

Answer:

Here is your answer

Step-by-step explanation:

A cubic spline is a spline constructed of piecewise third order polynomials which pass through a set of control points. The second derivative of each polynomial is commonly set to zero at the endpoints since this provides a boundary condition that completes the system of equations.

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Simplify (*x* +2x? - 5x)+(-3x + x +1)+(3+* + 2x).

Mama L [17]

Answer:

D

Step-by-step explanation:

4x^4+3x^2-3x^3-3x+1

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3 years ago

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Sam bought some toys. He bought a baseball for $5.88, and paid $8.41

myrzilka [38]

Answer:

$5.71

Step-by-step explanation:

20-5.88=14.12

14.12-8.41=5.71

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3 years ago

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If you have a cubic polynomial of the form y = ax^3 + bx^2 + cx + d and lets say it passes through the points (2,28), (-1, -5),

nikklg [1K]

Step-by-step explanation:

<u>Step 1: Solve using the first point</u>

<em>(2, 28)</em>

$28 = a(2)^3 + b(2)^2 + c(2) + d$

$28 = 8a + 4b + 2c + d$

<u>Step 2: Solve using the second point</u>

<em>(-1, -5)</em>

$-5 = a(-1)^3 + b(-1)^2 + c(-1) + d$

$-5 = -a + b - c + d$

<u>Step 3: Solve using the third point</u>

<em>(4, 220)</em>

$220 = a(4)^3 + b(4)^2 + c(4) + d$

$220 = 64a + 16b + 4c + d$

<u>Step 4: Solve using the fourth point</u>

<em>(-2, -20)</em>

$-20 = a(-2)^3 + b(-2)^2 + c(-2) + d$

$-20 = -8a + 4b - 2c + d$

<u>Step 5: Combine the first and fourth equations</u>

<u /> $28 - 20 = 8a - 8a + 4b + 4b + 2c - 2c + d + d$

$8 = 8b + 2d$

$8 - 8b = 8b - 8b + 2d$

$(8 -8b)/2 = 2d/2$

$4 - 4b = d$

<u>Step 6: Solve for c in the second equation</u>

$-5 + 5 = -a + b - c + d + 5$

$0 + c = -a + b - c + c + d + 5$

$c = -a + b + d + 5$

<u>Step 7: Substitute d with the stuff we got in step 5</u>

$c = -a + b + (4 - 4b) + 5$

$c = -a + b + 4 - 4b + 5$

$c = -a - 3b + 9$

<u>Step 8: Substitute d and c into the first equation</u>

<u /> $28 = 8a + 4b + 2(-a - 3b + 9) + (4 - 4b)$

$28 = 8a + 4b - 2a - 6b + 18 + 4 - 4b$

$28 - 22 = 6a - 6b + 22 - 22$

$6 / 6 = (6a - 6b) / 6$

$1 + b = a - b + b$

$1 + b = a$

<u>Step 9: Substitute a, b, and c into the third equation</u>

$220 = 64(1 + b) + 16b + 4(-(1 + b) - 3b + 9) + (4 - 4b)$

$220 = 64 + 64b + 16b + 4(-1 - b - 3b + 9) + 4 - 4b$

$220 - 100 = 60b + 100 - 100$

$120 / 60 = 60b / 60$

$2 = b$

<u>Step 10: Find a using b = 2</u>

$a = b + 1$

$a = (2) + 1$

$a = 3$

<u>Step 11: Find c using a = 3 and b = 2</u>

$c = -a - 3b + 9$

$c = -(3) - 3(2) + 9$

$c = -3 - 6 + 9$

$c = 0$

<u>Step 12: Find d using b = 2</u>

$d = 4 - 4b$

$d = 4 - 4(2)$

$d = 4 - 8$

$d = -4$

Answer: $a = 3, b = 2, c = 0,d = -4$

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3 years ago

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(2y-at)(y+2at) find the product

Fynjy0 [20]

(2y-at)(y+2at)

Answer: <span><span><span>−<span><span>2<span>a^2</span></span><span>t^2</span></span></span>+<span><span><span>3a</span>t</span>y</span></span>+<span>2<span>y^<span>2

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3 years ago

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Write the quadratic equation that has roots -1-rt2/3 and -1+rt2/3 if its coefficient with x^2 is equal to 1

weeeeeb [17]

The equation of the quadratic function is f(x) = x²+ 2/3x - 1/9

<h3>How to determine the quadratic equation?</h3>

From the question, the given parameters are:

Roots = (-1 - √2)/3 and (-1 + √2)/3

The quadratic equation is then calculated as

f(x) = The products of (x - roots)

Substitute the known values in the above equation

So, we have the following equation

$f(x) = (x - \frac{-1-\sqrt{2}}{3})(x - \frac{-1+\sqrt{2}}{3})$

This gives

$f(x) = (x + \frac{1+\sqrt{2}}{3})(x + \frac{1-\sqrt{2}}{3})$

Evaluate the products

$f(x) = (x^2 + \frac{1+\sqrt{2}}{3}x + \frac{1-\sqrt{2}}{3}x + (\frac{1-\sqrt{2}}{3})(\frac{1+\sqrt{2}}{3})$

Evaluate the like terms

$f(x) = x^2 + \frac{2}{3}x - \frac{1}{9}$

So, we have

f(x) = x²+ 2/3x - 1/9

Read more about quadratic equations at

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1 year ago