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3 years ago

12

Findtheequationoftheaxisofsymmetryfortheparabolay = x2 + 4. Simplify any numbers and write them as proper fractions, improper fr

actions, or integers.

2 answers:

Misha Larkins [42]3 years ago

4 0

Us 2(x+2)

Factor out from the expression

Alina [70]3 years ago

4 0

Answer:

x= 0

Step-by-step explanation:

I'm right

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Which expression is equivalent to 3x + 5x( x - 2)

In-s [12.5K]

I think it is c but pet me know

7 0

3 years ago

Find the difference (-6d+2)-(7+2d)

Cerrena [4.2K]

Answer:

-8d-5

HOPE THIS HELPS

- Todo ❤️

Step-by-step explanation:

-6d-2d=-8d

2-7=-5

6 0

3 years ago

Read 2 more answers

Lyrx [107]

Answer:

The years are:

1000 BCE, 999 BCE, 888 BCE, 777 BCE, 666 BCE, 555BCE, 444 BCE. 333 BCE, 222 BCE, 111 BCE

111 CE, 222 CE, 333 CE, 444 CE, 555 CE, 666 CE, 777 CE, 888 CE, 999 CE, 1000CE, 1011 CE, 1101 CE, 1110 CE,1222CE, 1333CE, 1444CE, 1555 CE, 1666 CE, 1777CE, 1888 CE, 1999CE, and 2000 CE

Explanation:

<u>1. Years BC:</u>

a) Years with four digits:

The first number with 3 equal digits is 1000. After that the years go decreasing: 999, 998, 997, ...

b) Years with three digits:

From 999 to 111, the numbers have three digits, thus the only that are solutions ara 999, 888, 777, 666, 555, 444, 333, 222, and 111: 9 numbers

After that the years have two digits, thus no solutions, with two digits.

Hence, we count 10 different years.

<u>2. Years CE</u>

a) Years with three digits:

111, 222, 333, 444, 555, 666, 777, 888, 999: 9 years

b) Years with four digits

i) Starting with 1:

With three 0: 1000: 1 year

With three 1: 1011, 1101, 1110: 3 years

With three digits different to 1: 1222, 1333, 1444, 1555, 1666, 1777, 1888, 1999: 8 years

ii) Starting with 2:

With three 0: 2000: 1 year

The next one with three equal digits is 2111 and it is after 2020 CE.

Therefore, 9 + 1 + 3 + 8 + 1 = 22 years starting with 2.

<u>3. Total</u>

<u />

10 years BC and 22 years CE have exactly three digits the same: 10 + 22 = 32.

3 0

4 years ago

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

What is the value of y

tangare [24]

Y=14.5
X= 5
Hope this helps :)

3 0

4 years ago