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3 years ago

I don’t know what to do please help

Mathematics

2 answers:

GalinKa [24]3 years ago

7 0

Answer:

haha um make conversation

Step-by-step explanation:

Snezhnost [94]3 years ago

6 0

Ok from experience texting back multiple times when the person is just leaving you on scene is a little weird. There could be multiple things happening and constantly trying to get a answer is kinda pressuring the person to text back even if they don’t know what to say.

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Student tickets to the school football game cost $5.80 each and general admission

Tems11 [23]

Answer:

x + y = 380

5.8x + 7.4y = 2372

Step-by-step explanation:

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3 years ago

A middle school has 419 students and 22 homerooms. How many students are in each homeroom and how many students are left over?

Murrr4er [49]

There are nineteen students in each homeroom, and three people left over.

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3 years ago

Read 2 more answers

Tell if the following statements are true or false :

azamat

Answer:

A.false, B.true

Step-by-step explanation:

3 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