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4 years ago

Guys, I don't understand this question is this correct??

Mathematics

2 answers:

poizon [28]4 years ago

8 0

Answer:

Step-by-step explanation:

I agree with your answer. The question is basically just saying the beverage company randomly chooses one of those letters to put under the bottle caps. For example, one bottle cap might have a T on it, and another might have an A on it. If you try to get all 4 letters, then you will have to keep buying those drinks until you manage to collect all 4 letters. Therefore, there are only 4 possible options for which letter you can get.

If you spin a spinner that has 4 equal sections, then this is similar. You have 4 possible options you can get.

Hope this helps! :)

Gennadij [26K]4 years ago

3 0

Answer:

I think you are correct. bc you have 4 different caps, and four different sections of a spinner that are equal same with the cap. So I believe so.

Step-by-step explanation:

I hope this helps! Im in algebra 2 so its been awhile since i have done this. Good Luck!!:)

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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.