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Vsevolod [243]
3 years ago
13

Help me pls ASAP I need it in 5mins​

Mathematics
2 answers:
love history [14]3 years ago
6 0

First option, empty set

iogann1982 [59]3 years ago
6 0
It’s empty set I believe
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An iPad for $480 and paid $31.20 in sales tax. What was the sales tax<br> rate?
Alborosie

Answer:

6.5%

Step-by-step explanation:

If you take 31.20 and divide it by 480, you get .065, which would then be converted into percentages by multiplying it by 100 (6.5%).

8 0
4 years ago
Can you guys help me please im so bad at math Lol so ima be posting more den one
yan [13]

Answer:

9

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
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.

We can summarize this process in the following table.

The approximation x_{5} gives you the root of the equation.

When you plot the equation you find that only have one real root and is approximate to the value found.

5 0
3 years ago
58.21 + ( -139.58 ) Do I do KCO?
Daniel [21]

Answer:

2136

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
algebra 2 which of the following sets is not finite? a. {x | x is a natural number less than 10} b. {x | x is a whole number bet
Luda [366]
A=\{0;\ 1;\ 2;\ 3;\ 4;\ 5;\ 6;\ 7;\ 8;\ 9\}-finite\\\\B=\{1;\ 2;\ 3;\ 4;\ 5;\ 6;\ 7;\ 8;\ 9\}-finite\\\\C=\{...;-4;-3;-2;-1\}-not\ finite
4 0
3 years ago
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