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3 years ago

If x < 0 and y < 0, where is the point (x, y) located?

Mathematics

1 answer:

Ipatiy [6.2K]3 years ago

4 0

Answer:

Quadrant III

Step-by-step explanation:

Quadrant I -> (+x,+y)

Quadrant II -> (-x,+y)

Quadrant III -> (-x,-y)

Quadrant IV -> (+x,-y)

If x and y must both be less than 0, then they must both be negative. Therefore, Quadrant III is your answer.

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The second one is bigger as the square is a even number with the negative whole

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3 years ago

Determine the values of the constants b and c so that the function given below is differentiable. f(x)={2xbx2+cxx≤1x>1

Lera25 [3.4K]

Assuming the function is

$f(x)=\begin{cases}2x&\text{for }x\le1\\bx^2+cx&\text{for }x>1\end{cases}$

For $f(x)$ to be differentiable, it necessarily has to be continuous. For this condition to be met, we need

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$
$\iff\displaystyle\lim_{x\to1}2x=\lim_{x\to1}(bx^2+cx)$
$\iff2=b+c$

For the derivative to exist, the one-sided limits of the derivative must also exist and be equal. We have

$f'(x)=\begin{cases}2&\text{for }x1\end{cases}$

$\displaystyle\lim_{x\to1^-}2=\lim_{x\to1^+}(2bx+c)$
$\iff2=2b+c$

Now we solve for $b$ and $c$ :

$\begin{cases}b+c=2\\2b+c=2\end{cases}\implies b=0,c=2$

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4 years ago

1.4.7+4.7.10+7.10.13 to n terms

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◆ SEQUENCE AND SERIES ◆

Heya !

Check the attachment.
Hope it helps you ;)

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3 years ago

The expression once The left side of an equation is shown below. 3(x+1)+9 if the equation has no solution which expression can b

lys-0071 [83]

<h3>Answer:</h3>

C. 4(x-3)-x

<h3>Step-by-step explanation:</h3>

All of the given expressions are equivalent to 3x+12 except selection C. Using that in your equation makes it be ...

... 3(x +1) +9 = 4(x -3) -x

... 3x +12 = 3x -12

... 12 = -12 . . . . . <em>false</em>

There is no value of x that will make this true, hence NO SOLUTION.

_____

<em>Comment on the other choices</em>

3x+12 = 3x+12 has an <em>infinite number of solutions</em>, as <u><em>any</em></u> value of x will make this true.

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3 years ago

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Need help please. I need help with how to do these problems.

Vikki [24]

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3 years ago