The domain is the set of x values such that $x \ge -\frac{1}{2}$ , basically x can be equal to -1/2 or it can be larger than -1/2. To get this answer, you solve $2x+1 \ge 0$ for x (subtract 1 from both sides; then divide both sides by 2). I set 2x+1 larger or equal to 0 because we want to avoid the stuff under the square root to be negative.

If you want the domain in interval notation, then it would be $\Big[ -\frac{1}{2} , \infty \Big)$ which means the interval starts at -1/2 (including -1/2) and then it stops at infinity. So technically it never stops and goes on forever to the right.

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Part b.

I'm going to use "sqrt" as shorthand for "square root"

f(x) = sqrt(2x+1)

f(10) = sqrt(2*10+1) ... every x replaced by 10

f(10) = sqrt(20+1)

f(10) = sqrt(21)

f(10) = 4.58257569 which is approximate

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Part c.

f(x) = sqrt(2x+1)

f(x) = sqrt(2(x)+1)

f(x+2a) = sqrt(2(x+2a)+1) ... every x replaced by (x+2a)

f(x+2a) = sqrt(2x+4a+1) .... distribute

we can't simplify any further

6 0

3 years ago

Cal is measuring temperature changes in four substances over different time periods as part of a school project. Which data show

djyliett [7]

Answer:

C) The third table. K= 5

$C)\frac{10}{2}=\frac{15}{3}=\frac{25}{5}=\frac{40}{8}=k= 5$

Step-by-step explanation:

1) Below there are the missing data:

A proportional relationship through a constant k. It is obtained when we divide:

$k=\frac{y}{x}$

2)In this case, when we divide the temperature (T) by the time (h).

$k=\frac{T}{h}$

3)So, examining the table below we are searching for a ratio k common to all measures (temperatures over hour).

$A) \frac{12}{3}\neq\frac{25}{5}\neq\frac{36}{6}\neq\frac{81}{9}\\B) \frac{22.5}{3}\neq\frac{30}{4}\neq\frac{52.5}{7}\neq\frac{67.5}{8}\\C)\frac{10}{2}=\frac{15}{3}=\frac{25}{5}=\frac{40}{8}=k= 5\\D)\frac{8.5}{2}\neq\frac{22.5}{5}\neq\frac{28.5}{7}\neq\frac{36}{8}$

3 0

3 years ago