The angle = 67<span>° and the pole is a right-angled triangle. Therefore the distance required to find (say "x) is the angle opposite the right-angle. The following equation needs to be solved:

cos 67 = 137 / x

Solving for x;
x = 350.62 ft</span>

3 0

3 years ago

3) g(x)= x3 + x<br> h(x) = x + 4<br> Find g(0)h(0)

Arte-miy333 [17]

Answer:

$goh(x)=x^3+12x^2+49x+68$

Step-by-step explanation:

$g(x) = x^3+x$

$h(x)=x+4$

We have to find

$goh(x)$

$goh(x) = g(h(x))=g(x+4)$

If $g(x) = x^3+x$

$g(x+4) = (x+4)^3+(x+4)$

= $x^3+12x^2+49x+68$

Hence our answer is

$goh(x)=x^3+12x^2+49x+68$

4 0

3 years ago

According to data from a medical association, the rate of change in the number of hospital outpatient visits, in millions, in

GaryK [48]

Answer:

a) $f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

$f'(t)=0.001155t(t-1980)^{0.5}$

a) To find the function f(t) you integrate f(t):

$\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt$

To solve the integral you use:

$\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}$

Next, you replace in the integral:

$\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C$

Then, the function f(t) is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'$

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.

Hence C' = 264,034,000

The function is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) For t = 2015 you have:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

3 0

3 years ago