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3 years ago

What is 4 + 4 PLZ HELP MEE

Mathematics

2 answers:

Alchen [17]3 years ago

8 0

Answer:

4 +4 = 8

Step-by-step explanation:

4+4=8

8 is the answer

Elenna [48]3 years ago

8 0

Answer:

Step-by-step explanation:

1 + 1 + 1 + 1 = 4

1 + 1 + 1 +1 = 4

4 + 4 = 8

You might be interested in

PLEASE SIMPLIFY THIS<br>

AnnyKZ [126]

Answer:

15y / 8

Step-by-step explanation:

Multiplying by 2 to get the common denominator.

3y / 4 * 2 = 6y / 8

Combine like terms.

6y / 8 + 9y / 8

Final Answer:

15y / 8

3 0

3 years ago

ASAP! GIVING BRAINLIEST! Please read the question THEN answer CORRECTLY! NO guessing. Show your work OR give an explaination.

erma4kov [3.2K]

Answer:

y = x² - 8x + 17

Step-by-step explanation:

Recall this formula:

y = a(x-h)² + k

Original equation was y = x², which in this form looks like

y = 1(x - 0)² + 0

If the shape of the graph didn't change, that means that a (the compressor/intensifier) stayed the same (a - value) meaning we can write the equation as..

y = (x-h)² + k

Remember, (h, k) is the vertex, which looking from your graph is at (4, 1)

y = (x - 4)² + 1

y = x^2- 8x + 16+1

y = x² - 8x + 17

8 0

4 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