All categories

3 years ago

If f (x) = StartRoot one-half x minus 10 EndRoot + 3, which inequality can be used to find the domain of f(x)?

Mathematics

2 answers:

strojnjashka [21]3 years ago

7 0

$\left(\frac{1}{2}\right) x-10+3 \geq 0$ is the inequality that can be used to find the domain of given f(x).

Step-by-step explanation:

For a square root function given by $f(x)=\sqrt{a(x)}$ , to have real values, the radicand x must be positive or equal to zero. So, domain for f(x) would be,

$a x \geq 0$

Given:

$f(x)=\sqrt{\left(\frac{1}{2}\right) x-10}+3$

In the given expression, under the square root, in place of ‘x’ presents as below and so

$\left(\frac{1}{2}\right) x-10+3 \geq 0$

The inequality depends on the true form of the given term, so it should be,

$\left(\frac{1}{2}\right) x-10+3 \geq 0$

velikii [3]3 years ago

6 0

Answer:

the answer is C

Step-by-step explanation:

have a great day

You might be interested in

PLEASE HELP ME I DONT WANT TO FAIL

Ksenya-84 [330]

Answer:

294

Step-by-step explanation:

6 0

3 years ago

One small tank takes twice as long to fill a tank as a larger pipe does. If it takes the two pipes 40 minutes to fill the tank w

Gala2k [10]

Um... it'll take each pipe 80 mins, bcz if together they take that long, then to find the pipes time separately, you just multiply it by two to show it doing the work the other pipe did as well. Is that what you wanted??

8 0

3 years ago

There are 63 students marching in a band, and they’re marching in 7 rows. How many students are in each row?

VLD [36.1K]

Total divided by number of rows=number per row
63/7=9
9 per row

4 0

3 years ago

Read 2 more answers

What expressions is equivalent to −2(x−5)?

Ksenya-84 [330]

Answer:

-2x + 10

Step-by-step explanation:

Distribute -2 to x - 5 to simplify the expression. So we get: -2(x - 5) = -2x + 10.

6 0

2 years ago

Sqaure root of 5 (u-1)(u^5+u^4+u^3+u^2+u+1)

Anna71 [15]

GIVEN:

$5(u - 1)( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1)$

remember:

$\sqrt{u} = {u}^{ \frac{1}{2} }$

And

${u}^{n} \times {u}^{m} = {u}^{n + m}$

SOLVE:

start by multiplying the factors:

$5( ({u}^{6} + {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u ) - ( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1))$

simplify by combing like terms. Most terms subtract off, leaving:

$5( {u}^{6} - 1)$

This can be factored, but it is not a perfect square, which is really what we need to take the square root.

$5( {u}^{3} - 1)( {u}^{3} + 1)$

I'm not exactly sure what form they want the answer in...

so taking the square root:

$\sqrt{5( {u}^{6} - 1) } = {(5( {u}^{6} - 1))}^{ \frac{1}{2} }$

so my best answer is:

${5}^{ \frac{1}{2} } \times {( {u}^{6} - 1)}^{ \frac{1}{2} }$
or the more factored form:

${5}^{ \frac{1}{2} } { ({u}^{3} - 1)}^{ \frac{1}{2} } { ({u}^{3} + 1 )}^{ \frac{1}{2} }$

I'm not sure how else to solve it. Taking the square root doesn't work out super well, so I left it in the most simple form I could.

sorry for not coming to a definitive answer!

7 0

3 years ago