All categories

3 years ago

25h + 25 = c 15h + 40 = c

Mathematics

1 answer:

GuDViN [60]3 years ago

8 0

Answer:

c= 125/2

h= 3 /2

Step-by-step explanation:

You might be interested in

Find the average rate of change for the function, f(x) = -3x2 + 2x - 5 over the interval -3 < x < 3.

aev [14]

Answer:

Step-by-step explanation:

4 0

2 years ago

-6b + 20 = -16<br> what is b?<br> (-6 x __) + 20 = -16

denpristay [2]

-6b = -36
Divide by -6
B is 6

6 0

3 years ago

How to graph the function f(x)=-2/3x+490

julia-pushkina [17]

Start on the y-axis and put a point at 490 bc it the y-intercept and then from there go up 2 and right 3 and put a point there and keep going and to go down you just go down 2 and left 3 and the point it.

7 0

3 years ago

Read 2 more answers

An airplane climbs at an angle of 2.7" from an altitude of 4500 ft to an altitude of 5600 ft. How far does the airplane travel a

Dvinal [7]

Answer:

23,325 ft, about 4.42 miles

Step-by-step explanation:

The geometry of the situation can be modeled by a right triangle, where the side adjacent to the 2.7° angle of climb is the horizontal distance, and the side opposite is the vertical change in altitude. That change is ...

5600 ft -4500 ft = 1100 ft

and the angle relation is ...

tan(2.7°) = opposite/adjacent = (1100 ft)/(horizontal distance)

Multiplying this equation by (horizontal distance)/tan(2.7°) gives ...

horizontal distance = (1100 ft)/tan(2.7°) ≈ 23,325 ft

Dividing this by 5280 ft/mi gives ...

horizontal distance ≈ 4.42 mi

The airplane travels about 23,325 ft, or 4.42 miles, horizontally as it climbs.

4 0

3 years ago

Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x

soldier1979 [14.2K]

Answer:

(a) The function is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is $x = -\frac{1}{2}$

(d) The function is concave upward on $\left(- \frac{1}{2}, \infty\right)$ and concave downward on $\left(-\infty, - \frac{1}{2}\right)$

Step-by-step explanation:

(a) To find the intervals where $f(x) = 2x^3 + 3x^2 -180x$ is increasing or decreasing you must:

1. Differentiate the function

$\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180$

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

$f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6$

These points divide the number line into three intervals:

$(-\infty,-6)$ , $(-6,5)$ , and $(5, \infty)$

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

$\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right$

Therefore f(x) is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.

f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

$f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6$

We set f''(x) = 0

$f''(x) =12x+6 =0\\\\x=-\frac{1}{2}$

Analyzing concavity, we get

$\left\begin{array}{cccc}Interval&x-value&f''(x)\\(-\infty,-1/2)&-2&-18\\(-1/2,\infty)&0&6\\\end{array}\right$

The function is concave upward on $(-1/2,\infty)$ because the f''(x) > 0 and concave downward on $(-\infty,-1/2)$ because the f''(x) < 0.

f(x) is concave down before $x = -\frac{1}{2}$ , concave up after it. So f(x) has an inflection point at $x = -\frac{1}{2}$ .

7 0

3 years ago