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3 years ago

Explain natural substance

Mathematics

2 answers:

sammy [17]3 years ago

7 0

Answer:

This is your answer ☺️

Luden [163]3 years ago

4 0

Biotic materials. Wood: rattan, bamboo, bark
Natural fiber: silk, wool, cotton, flax, hemp, jute, kapok, kenaf, moss

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2b) A farmer is building a pen inside a barn. The pen will be in the shape of a right triangle.

kenny6666 [7]

Answer:

Step-by-step explanation:

the sum of any two sides of a right triangle more than any other side

14+15>29, x is less than 29

14+x>15

x>1

15+x>14

x>-1

1<x<29

lets use Pythagorean Theorem

14^(2)+15^(2)=c^2

196+225=c^2

421=c^2

plusminus sqrt(421)=c

distance can't be negative, so:

c=sqrt(421)

c=20.5182845287

sqrt(421) is the largest possible right triangle side

1<x $\leq$ sqrt(421)

B) as we can see from above, the largest possible length of the third side is 14/sqrt(421) or about 20.5182845287ft

C)sqrt(421) by 14 by 15

tan(x1)=15/14

x1=tan^-1(15/14)

x1=47 degrees

tan(x2)=14/15

x2=tan^-1(14/15)

x2=43 degrees

7 0

3 years ago

Find the roots of the equation f(x) = x3 - 0.2589x2 + 0.02262x -0.001122 = 0

devlian [24]

Answer:

The root of the equation $x^3-0.2589x^{2}+0.02262x-0.001122=0$ is x ≈ 0.162035

Step-by-step explanation:

To find the roots of the equation $x^3-0.2589x^{2}+0.02262x-0.001122=0$ you can use the Newton-Raphson method.

It is a way to find a good approximation for the root of a real-valued function f(x) = 0. The method starts with a function f(x) defined over the real numbers, the function derivative f', and an initial guess $x_{0}$ for a root of the function. It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it.

This is the expression that we need to use

$x_{n+1}=x_{n} -\frac{f(x_{n})}{f(x_{n})'}$

For the information given:

$f(x) = x^3-0.2589x^{2}+0.02262x-0.001122=0\\f(x)'=3x^2-0.5178x+0.02262$

For the initial value $x_{0}$ you can choose $x_{0}=0$ although you can choose any value that you want.

So for approximation $x_{1}$

$x_{1}=x_{0}-\frac{f(x_{0})}{f(x_{0})'} \\x_{1}=0-\frac{0^3-0.2589\cdot0^2+0.02262\cdot 0-0.001122}{3\cdot 0^2-0.5178\cdot 0+0.02262} \\x_{1}=0.0496021$

Next, with $x_{1}=0.0496021$ you put it into the equation

$f(0.0496021)=(0.0496021)^3-0.2589\cdot (0.0496021)^2+0.02262\cdot 0.0496021-0.001122 = -0.0005150$ , you can see that this value is close to 0 but we need to refine our solution.

For approximation $x_{2}$

$x_{2}=x_{1}-\frac{f(x_{1})}{f(x_{1})'} \\x_{1}=0-\frac{0.0496021^3-0.2589\cdot 0.0496021^2+0.02262\cdot 0.0496021-0.001122}{3\cdot 0.0496021^2-0.5178\cdot 0.0496021+0.02262} \\x_{1}=0.168883$

Again we put $x_{2}=0.168883$ into the equation

$f(0.168883)=(0.168883)^3-0.2589\cdot (0.168883)^2+0.02262\cdot 0.168883-0.001122=0.0001307$ this value is close to 0 but again we need to refine our solution.