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3 years ago

Can you always draw a piecewise function without picking up your pencil.

Mathematics

2 answers:

MArishka [77]3 years ago

5 0

Answer:

no?

Step-by-step explanation:

lisov135 [29]3 years ago

4 0

Answer:

Yes

Step-by-step explanation:

A function is continuous on an interval if the graph has no breaks, jumps, or holes. You can draw it without picking up your pencil.

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PLZ ANSWER ASAP! 15 points

Mariulka [41]

Answer: THe last one.

Step-by-step explanation:

This question is, in essence, basically asking which numbers are less than -3. Looking at the numbers, it is clearly the last set.

5 0

3 years ago

Can someone please answer this, ill give you brainliest Would be very appreciated.

Svetllana [295]

Explanation:

f(x) = (x-4)(x+2)

1) For x-intercept, y will be 0

(x-4)(x+2) = 0

(x-4) =0, (x+2) = 0

x = 4, x = -2

x-intercept: (4, 0), (-2, 0)

2) For vertex: x = -b/2a where ax² + bx + c

Quadratic function:

(x-4)(x+2)

x² +2x-4x -8

x² -2x -8

vertex:

x = -(-2)/2(1)

x = 1

y: (x-4)(x+2) = (1-4)(1+2) = -9

ordered pair of vertex: (1, -9)

3) For y-intercept, x will be 0

(x-4)(x+2)

(0-4)(0+2)

-8

y-intercept: (0, -8)

8 0

2 years ago

Read 2 more answers

What is 5(x-2)-14x when x=4

nevsk [136]

Answer:

-46

Step-by-step explanation:

5 (x - 2) - 14x. x=4

5 (4 - 2) - 14(4) (distribute)

20 - 10 - 14(4) (multiply)

20 - 10 - 56 (subtract and solve.)

10 - 56 = -46 - solution

hope this helps (:

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20-%20x%20%5E%7B3%7D%20%7D%20" id="TexFormula1" title=" \sqrt{ - x ^{3} } " alt=

noname [10]

I'm guessing you're given the function $y(x)=2-x^3$ , and you're asked to find the inverse function $y^{-1}(x)$ . To do this, swap $x$ and $y$ , then solve for $y$ :

$x=2-y^3\implies y^3=2-x\implies y=(2-x)^{1/3}=\sqrt[3]{2-x}$

so that the inverse function is

$y^{-1}(x)=\sqrt[3]{2-x}$

Just to verify:

$y(y^{-1}(x))=y(\sqrt[3]{2-x})=2-(\sqrt[3]{2-x})^3=2-(2-x)=x$

$y^{-1}(y(x))=y^{-1}(2-x^3)=\sqrt[3]{2-(2-x^3)}=\sqrt[3]{x^3}=x$

But in case you're actually only interested in computing the square root, first we note that $\sqrt x$ (the real-valued square root) is only defined as long as $x\ge0$ . So $\sqrt{-x^3}$ is defined as long as $-x^3\ge0$ , or $x^3\le0$ , or equivalently $x\le0$ . Under this condition, we could write

$\sqrt{-x^3}=\sqrt{-x\times x^2}=\sqrt{-x}\sqrt{x^2}$

We can simplify this further, but we have to be careful. Suppose $x=-1$ . Then $x^2=(-1)^2=1$ . But we get the same result if $x=1$ , since $x^2=1^2=1$ . There are two possible values of $x$ that given the same value of $x^2$ , so to capture both of them, we take $\sqrt{x^2}=|x|$ , the absolute value of $x$ . Then