7 divided by somethings equal 7.35

1 answer:

6 0

Let's translate it into an equation and then solve...

7/x = 7.35
x * 7/x = 7.35 * x
7 = 7.35x
7/7.35 = x
.9524 = x

So 7 divided by .9524 (rounded to the nearest ten thousandth) equals 7.35

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cupoosta [38]

2. $f(x)=\dfrac18\left(\dfrac14\right)^{x-2}$

Domain: $(-\infty,\infty)$ , because any value of $x$ is allowed and gives a number $f(x)$ .

Range: $(0,\infty)$ , because $a^x>0$ for any positive real $a\neq0$ .

y-intercept: This is a point of the form $(0, f(0))$ . So plug in $x=0$ ; we get $f(0)=\frac18\left(\frac14\right)^{-2}=\frac{4^2}8=2$ . So the intercept is (0, 2), or just 2. (Interestingly, you didn't get marked wrong for that...)

Asymptote: This can be deduced from the range; the asymptote is the line $y=0$ .

Increasing interval: Going from left to right, there is no interval on which $f(x)$ is increasing, since 1/4 is between 0 and 1.

Decreasing interval: Same as the domain; $f(x)$ is decreasing over the entire real line.

End behavior: The range tells you $f(x)>0$ , and you know $f(x)$ is decreasing over its entire domain. This means that $f(x)\to+\infty$ as $x\to-\infty$ , and $f(x)\to0$ and $x\to+\infty$ .

3. $f(x)=\left(\dfrac32\right)^{-x}-7$

Domain: Same as (2), $(-\infty, \infty)$ .

Range: We can rewrite $f(x)=\left(\frac23\right)^x-7$ . $\left(\frac23\right)^x>0$ for all $x$ , so $\left(\frac23\right)^x-7>-7$ for all $x$ . Then the range is $(-7,\infty)$ .