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3 years ago

During each of the past three months, Tyler grew

1 answer:

nataly862011 [7]3 years ago

4 0

First you have to make all the fractions have a common denominator to add them so since 3/16 is the largest denominator you have to convert the others to have the denominator of 16. 1/8 = 2/16. 1/4 = 4/16. so if you add the numerators that’s 3 + 2 + 4 = 9. so the answer would be 9/16, and this cannot be simplified so it is your final answer:)

You might be interested in

Find the radius of a circle with a circumference of 35 pie yards

natita [175]

Answer:

Using the formula

C=2πr

Solving for r

r=C

2π=35

2·π≈5.57042

Step-by-step explanation:

8 0

2 years ago

Consider the equation below. X2 + y2 − 2x − 4y − z + 5 = 0 reduce the equation to one of the standard forms. Classify the surfac

jeka57 [31]

we are given

$x^2+y^2-2x-4y-z+5=0$

standard form:

We will complete x , y and z square

make all x , y and z terms together

$x^2-2x+y^2-4y-z+5=0$

$x^2-2x+y^2-4y-z=-5$

$x^2-2*1*x+y^2-2*2*y-z=-5$

$x^2-2*1*x+1^2+y^2-2*2*y+2^2-z=-5+1^2+2^2$

$(x-1)^2+(y-2)^2-z=0$

$z=(x-1)^2+(y-2)^2$

surface:

It is an infinite paraboloid

graph:

3 0

3 years ago

Do you have to multiply by 2 on circumference

OverLord2011 [107]

Circumference of a circle = 2π r (2 * π * radius) or πd (π * diameter)

3 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 $\frac{1}{3^3}= \frac{1}{27}$