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3 years ago

8

Ben is 20 years older than Daniel. Ben and Daniel first met two years ago. Three years ago, Ben was 3 times as old as Daniel. Ho

w old is Ben now?

1 answer:

serg [7]3 years ago

7 0

Daniel was 10 three years ago, and Ben was 30. Now, Ben is 33.

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(3x + 20) (5x) find x

HACTEHA [7]

15x^2+100x

hope this helps

7 0

2 years ago

A hawk soars at an altitude of 1,800ft. If the hawk descends to the ground in 45 min, what is the vertical speed?

melomori [17]

1800ft/45mins. I'm not sure.

6 0

3 years ago

Read 2 more answers

A. Use composition to prove whether or not the functions are inverses of each other. B. Express the domain of the compositions u

Kryger [21]

Given: $f(x) = \frac{1}{x-2}$

$g(x) = \frac{2x+1}{x}$

A.)Consider

$f(g(x))= f(\frac{2x+1}{x} )$

$f(\frac{2x+1}{x} )=\frac{1}{(\frac{2x+1}{x})-2}$

$f(\frac{2x+1}{x} )=\frac{1}{\frac{2x+1-2x}{x}}$

$f(\frac{2x+1}{x} )=\frac{x}{1}$

$f(\frac{2x+1}{x} )=1$

Also,

$g(f(x))= g(\frac{1}{x-2} )$

$g(\frac{1}{x-2} )= \frac{2(\frac{1}{x-2}) +1 }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{\frac{2+x-2}{x-2} }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{x }{1}$

$g(\frac{1}{x-2} )= x$

Since, $f(g(x))=g(f(x))=x$

Therefore, both functions are inverses of each other.

B.

For the Composition function $f(g(x)) = f(\frac{2x+1}{x} )=x$

Since, the function $f(g(x))$ is not defined for $x=0$ .

Therefore, the domain is $(-\infty,0)\cup(0,\infty)$

For the Composition function $g(f(x)) =g(\frac{1}{x-2} )=x$

Since, the function $g(f(x))$ is not defined for $x=2$ .

Therefore, the domain is $(-\infty,2)\cup(2,\infty)$

8 0

3 years ago

Simplify the rational expression. State any restrictions on the variable.

ExtremeBDS [4]

we are given

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}$

Firstly, we will factor numerators and denominators

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-6)(n^2-4)}{(n^2-3)(n^2-6)}$

we can see that

n^2 -6 is factor on both numerator and denominator

so, it will get cancelled

and n^2 -6 can not be equal to 0

so, one of restriction is

$n^2-6\neq 0$

$n\neq -+ \sqrt{6}$

we can simplify it

$\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-4)}{(n^2-3)}$

we know that denominator can not be zero

$n^2-3\neq 0$

$n\neq -+ \sqrt{3}$

so, option-B.......Answer

3 0

3 years ago

From a random sample of 58 businesses, it is found that the mean time the owner spends on administrative issues each week is 21.

zzz [600]

Answer: (20.86, 22.52)

Step-by-step explanation:

Formula to find the confidence interval for population mean :-

$\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}$

, where $\overline{x}$ = sample mean.

z*= critical z-value

n= sample size.

$\sigma$ = Population standard deviation.

By considering the given question , we have

$\overline{x}= 21.69$

$\sigma=3.23$

n= 58

Using z-table, the critical z-value for 95% confidence = z* = 1.96

Then, 95% confidence interval for the amount of time spent on administrative issues will be :

$21.69\pm (1.96)\dfrac{3.23}{\sqrt{58}}$

$=21.69\pm (1.96)\dfrac{1.7}{7.61577}$

$=21.69\pm (1.96)(0.223221)$

$\approx21.69\pm0.83$

$=(21.69-0.83,\ 21.69+0.83)=(20.86,\ 22.52)$

Hence, the 95% confidence interval for the amount of time spent on administrative issues = (20.86, 22.52)

7 0

3 years ago