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Gennadij [26K]
2 years ago
6

Work out 1/3 + 2/9 Give your answer in its simplest form

Mathematics
2 answers:
Soloha48 [4]2 years ago
4 0

Step-by-step explanation:

1/3 + 2/9 = 3/9 + 2/9 = 5/9

Topic: Fractions

If you like to venture further, feel free to check out my insta (learntionary). I'll be constantly posting math tips and notes! Thanks!

monitta2 years ago
3 0

Answer:

5/9

Step-by-step explanation:

You might be interested in
What is the answer to this queation for my maths homework<br> 3a+7=13
borishaifa [10]
3a+7=13
3a+7-7=13-7
3a=6
\frac{3a}{3}=\frac{6}{3}
a=2
5 0
3 years ago
Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d
andrey2020 [161]

Answer:

The maximum value is 1/27 and the minimum value is 0.

Step-by-step explanation:

Note that the given function is equal to (xyz)^2 then it means that it is positive i.e f(x,y,z)\geq 0.

Consider the function F(x,y,z,\lambda)=x^2y^2z^2-\lambda (x^2+y^2+z^2-1)

We want that the gradient of this function  is equal to zero. That is (the calculations in between are omitted)

\frac{\partial F}{\partial x} = 2x(y^2z^2 - \lambda)=0

\frac{\partial F}{\partial y} = 2y(x^2z^2 - \lambda)=0

\frac{\partial F}{\partial z} = 2z(x^2y^2 - \lambda)=0

\frac{\partial F}{\partial \lambda} = (x^2+y^2+z^2-1)=0

Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.

If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.

Suppose that x is zero. Then, from the second and third equation we have that

-2y\lambda = -2z\lambda. If lambda is not zero, then y =z. But, since -2y\lambda=0 and lambda is not zero, this implies that x=y=z=0 which is not possible. This proofs that if one of the variables is 0, then lambda is zero. So, having one or two variables equal to zero are feasible solutions for the problem.

Suppose that only x is zero, then we have the solution set y^2+z^2=1.

If both x,y are zero, then we have the solution set z^2=1. We can find the different solution sets by choosing the variables that are set to zero.

NOw, suppose that none of the variables are zero.

From the first and second equation we have that

\lambda = y^2z^2 = x^2z^2 which implies x^2=y^2

Also, from the first and third equation we have that

\lambda = y^2z^2 = x^2y^2 which implies x^2=z^2

So, in this case, replacing this in the restriction we have 3z^2=1, which gives as another solution set. On this set, we have x^2=y^2=z^2=\frac{1}{3}. Over this solution set, we have that the value of our function is \frac{1}{3^3}= \frac{1}{27}

4 0
3 years ago
Can you give me a example on how to do this that would be really helpfull
Rom4ik [11]
We'll do the first one.  I can't read the variable, but it looks like a \theta so that's what I'll use, but it doesn't matter.

First, get the variable by itself:

\theta

You get that before you multiply each side by two.  

 Now, you color the line so that all values that are less than 6 are shaded, so you put an open circle at 6 since it's not included and then you shade everything to the left.
6 0
3 years ago
Two events, X and Y, are independent of each other. P(Y) = 5/6 and P(X and Y) = 1/3. What is P(X) written as a decimal? Round to
Ksenya-84 [330]
For independent events,
P(X &cap; Y) = P(X)* P(Y)
=>
1/3 = P(X)*(5/6)
solve for P(X) =>
P(X) = (1/3)*(6/5) = 2/5 = 0.4
5 0
2 years ago
Read 2 more answers
U xy, for u = 2, x = 9, and y = 6
Yuki888 [10]
First thing you gotta do is to sub in the numbers into the equation/expression
<span>uxy = (2)(9)(6)
</span>      = 108

* brackets means multiplying

Final answer is 108
8 0
3 years ago
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