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3 years ago

Please someone help!

Mathematics

2 answers:

9966 [12]3 years ago

4 0

Answer:

I think that is right :/

Step-by-step explanation:

$10 \frac{3}{4} + 9 \frac{2}{4} \\ \frac{43}{4} + \frac{34}{4} \\ \frac{77}{4} \\ or \: \: 19 \frac{1}{4}$

pochemuha3 years ago

3 0

Your answer is 19 5/4

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Which function represents the graph of h(x)=2|x+3|−1 after it is translated 2 units right?

harina [27]

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$\bf ~~~~~~\textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if } B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }\frac{ C}{ B}\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}$

$\bf ~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\ \bullet \textit{ vertical shift by } D\\ ~~~~~~if\ D\textit{ is negative, downwards}\\\\ ~~~~~~if\ D\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{ B}$

now with that template in mind

$\bf h(x)=\stackrel{A}{2}|\stackrel{B}{1}x+\stackrel{C}{3}|\stackrel{D}{-1}\qquad \qquad \stackrel{\textit{C=C-2 a translation to the right}}{h(x)=2|x\boxed{+3-2}|-1} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill h(x)=2|x+1|-1~\hfill$

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4 years ago

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Hellen is using 3.508 grams of copper for her science experiment. How many kilograms of copper did Hellen use for her science ex

mash [69]

Answer:

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Step-by-step explanation:

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3 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Mkey [24]

For this case, we have to:

By definition, we know:

The domain of $f (x) = \sqrt [3] {x}$ is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.

So, we have:

$y = \sqrt [3] {x-2}$ with $x = 0$ : $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x+2}$ with $x = 0:\ y = \sqrt [3] {2}$ is also defined.

$f (x) = \sqrt {x}$ has a domain from 0 to ∞.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .