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3 years ago

8

State the range of the numbers displayed in column x and column y. xy 15 2739

1 answer:

Masja [62]3 years ago

6 0

We can not really tell in this question as you dont know the equation that is being used for the domain and range relationship but overall one should know that:

The set of values of the independent variable(s) for which a function or relation is defined as the domain of a function. Typically, this is the set of x-values that give rise to real y-values.

The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain.

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Please help me out I need help so bad

wel

the answer is definitly C

5 0

3 years ago

Mr.russo exchanged 200 euros for 270 dollars how many dollar he get foe 300 euros

dimulka [17.4K]

We must fine the conversion for this, so we have to divide $270 by 200€ to get 1.35 Dollars for every Euro..
Now we multiply 300 by 1.35 to get 405 dollars

6 0

2 years ago

Read 2 more answers

Carlos earns $11 per hour at his job. Let x represent the number of hours he worked in May. Let y represent the number of hours

Firlakuza [10]

11x+11y

11(x+y)

Explanation

Step 1

Let

Carlos earns == 11 per hour

x represents the number of hours he worked in May

the,

the amount he earned in Mayis

$11\cdot x=11x$

y represent the number of hours he worked in June.

the amount he earned in June was

$11\cdot y=11y$

Step 2

the amount of money he earned is May and June is the sum of the values

$\begin{gathered} \text{total}= \\ 11x+11y \\ \text{if we factorize 11} \\ 11(x+y) \end{gathered}$

I hope this helps you

5 0

7 months ago

Find the maximum and minimum values of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤100. The maximum value of f(x,y) is:

ryzh [129]

First find the critical points of <em>f</em> :

$f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7$

$\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1$

$\dfrac{\partial f}{\partial y}=6y=0\implies y=0$

so the point (1, 0) is the only critical point, at which we have

$f(1,0)=-7$

Next check for critical points along the boundary, which can be found by converting to polar coordinates:

$f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t$

Find the critical points of <em>g</em> :

$\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0$

$\implies\sin t=0\text{ OR }1+5\cos t=0$

$\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi$

where <em>n</em> is any integer. We get 4 critical points in the interval [0, 2π) at

$t=0\implies f(10,0)=155$

$t=\cos^{-1}\left(-\dfrac15\right)\implies f(-2,4\sqrt6)=299$

$t=\pi\implies f(-10,0)=235$

$t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299$

So <em>f</em> has a minimum of -7 and a maximum of 299.

4 0

3 years ago

AURORKA [14]

9 is in parentheses therefore it's elevated above 3 that means you have to do operations on it first.
3 + (-9) = -6
The sum is negative.

4 0

3 years ago