Need help in maximum value

Please solve this question. cant understand

saul85 [17]

Answer:

x=150°

Step-by-step explanation:

A=180-60=120

B=180-60=120

C=45

D=360-120=240

E=45

The sum of the interiors of a polygon is (n-2)*180. Since this is a hexagon, (6-2)*180=4*180=720. SO..

B=720-120-120-45-240-45=150

Assuming the questions is asking to solve for x, the angle is 150°

8 0

3 years ago

11th grade geometry:

Aliun [14]

Answer:

The perimeter is 72 units and the area is 149 square units.

Step-by-step explanation:

$\triangle SBA$ has coordinates $S(15,-8),B(-2,21)$ and $A(0,0)$

Using the distance formula.........

Length of side $SB = \sqrt{(15+2)^2+(-8-21)^2}= \sqrt{17^2+(-29)^2}= \sqrt{1130}$

Length of side $BA= \sqrt{(-2)^2+(21)^2}= \sqrt{445}$

Length of side $AS =\sqrt{(15)^2+(-8)^2}=\sqrt{289}=17$

So, the perimeter of the triangle will be: $(SB+BA+AS)= \sqrt{1130}+ \sqrt{445}+17 =71.71... \approx 72$ units. (Rounded to the nearest unit)

The height of the triangle for the corresponding base $SB$ is 8.89 units.

Formula for the Area of triangle, $A= \frac{1}{2}(base\times height)$

So, the area of the $\triangle SBA$ will be: $\frac{1}{2}(\sqrt{1130}\times 8.89)= 149.42... \approx 149$ square units. (Rounded to the nearest unit)

3 0

3 years ago

Find the expansion of tan x about the point X = 0.

yan [13]

Answer:

$f(x) = x +\frac{1}{3}x^{3}+....$

Step-by-step explanation:

As per the question,

let us consider f(x) = tan(x).

We know that The Maclaurin series is given by:

$f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......$

So, differentiate the given function 3 times in order to find f'(x), f''(x) and f'''(x).

Therefore,

f'(x) = sec²x

f''(x) = 2 × sec(x) × sec(x)tan(x)

= 2 × sec²(x) × tan(x)

f'''(x) = 2 × 2 sec²(x) tan(x) tan(x) + 2 sec²(x) × sec²(x)

= 4sec²(x) tan²(x) + 2sec⁴(x)

= 6 sec⁴x - 4 sec² x

We then substitute x with 0, and find the values

f(0) = tan 0 = 0

f'(0) = sec²0 = 1

f''(0) = 2 × sec²(0) × tan(0) = 0

f'''(0) = 6 sec⁴0- 4 sec² 0 = 2

By putting all the values in the Maclaurin series, we get

$f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......$

$f(x) = 0 + \frac{1}{1}\cdot x+ \frac{0}{2}\cdot x^{2}+\frac{2}{6}\cdot x^{3}+......$

$f(x) = x +\frac{1}{3}x^{3}+....$

Therefore, the expansion of tan x at x = 0 is

$f(x) = x +\frac{1}{3}x^{3}+....$ .

8 0

3 years ago

Can someone help with geometry !! If m

dedylja [7]

I can try to help you if you type the rest of your question!

8 0

3 years ago

Simplify each ratio 1. $8 to 62 cents 2. 14 inches to 2 yards 3. 5 days to 40 hours 4. Ratio of dogs to cats at the kennel if th

Vika [28.1K]

Answer:9

Step-by-step explanation:

4 0

4 years ago