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3 years ago

Help please I really need it

Mathematics

1 answer:

KatRina [158]3 years ago

8 0

First write the equation 8 + 12 > x because according to the triangle property, the sum of any 2 sides have to be greater than the third side. This means that the difference of any two sides should be smaller than the third side.

So, the answer would be 4 < x < 20.

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The body of a great roadrunner is 16 in. Long and the tail is 8 in. Long. The total length of a great roadrunner is 4 times long

arlik [135]

Answer: Love bird is 4 inches long.

Step-by-step explanation:

Given: The body of a great roadrunner is 16 in. Long and the tail is 8 in.

Also, Total length of a great roadrunner is = 4 x (length of lovebird).

It implies, length of lovebird = $\dfrac{\text{length of lovebird}}{4}$

$=\dfrac{16}{4}=4\text{ inches}$

Hence, Love bird is 4 inches long.

8 0

3 years ago

The Hudson Valley Stadium Renegades has a capacity of maximum 4,505 people. During a the peak of its popularity, the tickets sol

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219 x 4505 = 986,595

6 0

4 years ago

Write 4/12 in lowest terms

kondor19780726 [428]

Answer: 1/3

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Use lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y = xyz; x^

Snezhnost [94]

I'm assuming the constraint involves some plus signs that aren't appearing for some reason, so that you're finding the extrema subject to $x^2+2y^2+3z^2=96$ .

Set $f(x,y,z)=xyz$ and $g(x,y,z)=x^2+2y^2+3z^2-96$ , so that the Lagrangian is

$L(x,y,z,\lambda)=xyz+\lambda(x^2+2y^2+3z^2-96)$

Take the partial derivatives and set them equal to zero.

$\begin{cases}L_x=yz+2\lambda x=0\\L_y=xz+4\lambda y=0\\L_z=xy+6\lambda z=0\\L_\lambda=x^2+2y^2+3z^2-96=0\end{cases}$

One way to find the possible critical points is to multiply the first three equations by the variable that is missing in the first term and dividing by 2. This gives

$\begin{cases}\dfrac{xyz}2+\lambda x^2=0\\\\\dfrac{xyz}2+2\lambda y^2=0\\\\\dfrac{xyz}2+3\lambda z^2=0\\\\x^2+2y^2+3y^2=96\end{cases}$

So by adding the first three equations together, you end up with

$\dfrac32xyz+\lambda(x^2+2y^2+3z^2)=0$

and the fourth equation allows you to write

$\dfrac32xyz+96\lambda=0\implies \dfrac{xyz}2=-32\lambda$

Now, substituting this into the first three equations in the most recent system yields

$\begin{cases}-32\lambda+\lambda x^2=0\\-32\lambda+2\lambda y^2=0\\-32\lambda+3\lambda z^2=0\end{cases}\implies\begin{cases}x=\pm4\sqrt2\\y=\pm4\\z=\pm4\sqrt{\dfrac23}\end{cases}$

So we found a grand total of 8 possible critical points. Evaluating $f(x,y,z)=xyz$ at each of these points, you find that $f(x,y,z)$ attains a maximum value of $\dfrac{128}{\sqrt3}$ whenever exactly none or two of the critical points' coordinates are negative (four cases of this), and a minimum value of $-\dfrac{128}{\sqrt3}$ whenever exactly one or all of the critical points' coordinates are negative.

6 0

3 years ago

During one day in a hardware store, 37 people came in and made a purchase, and 13 people looked around but didn’t purchase anyth

MA_775_DIABLO [31]

A= 37:13
B= 13:37
C=37:50
D=13:50

8 0

3 years ago