Ask question

Ask question

All categories

faltersainse [42]

3 years ago

14

Use the equation given below to find f '' (π/ 3) . f(x) = sec x

1 answer:

vova2212 [387]3 years ago

8 0

F'(x) = secxtanx
f''(x) = (secx)^3 + (secx)^2*(tanx)^2
Plug in $\pi$ / 3 and you get 8 + 12 = 20

Check me on that i did it in my head but it should be right

You might be interested in

Help me i swear ill give brainlist

Elan Coil [88]

The area of a circle is pi • radius^2

so
3.4^2 • 3.14 = 36.3
Good luck! The answer is C!

4 0

3 years ago

A rectangle's length to width ratio is 9:4. If its length is 14 cm more than its width, what is its area?

sergiy2304 [10]

The area of the rectangle should be 87.1cm^2

6 0

3 years ago

4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513

dimaraw [331]

Answer:

Part 4) $sin(\theta)=\frac{12}{13}$

Part 10) The angle of elevation is $40.36\°$

Part 11) The angle of depression is $78.61\°$

Part 12) $arcsin(0.5)=30\°$ or $arcsin(0.5)=150\°$

Part 13) $-45\°$ or $225\°$

Step-by-step explanation:

Part 4) we have that

$cos(\theta)=-\frac{5}{13}$

The angle theta lies in Quadrant II

so

The sine of angle theta is positive

Remember that

$sin^{2}(\theta)+ cos^{2}(\theta)=1$

substitute the given value

$sin^{2}(\theta)+(-\frac{5}{13})^{2}=1$

$sin^{2}(\theta)+(\frac{25}{169})=1$

$sin^{2}(\theta)=1-(\frac{25}{169})$

$sin^{2}(\theta)=(\frac{144}{169})$

$sin(\theta)=\frac{12}{13}$

Part 10)

Let

$\theta$ ----> angle of elevation

we know that

$tan(\theta)=\frac{85}{100}$ ----> opposite side angle theta divided by adjacent side angle theta

$\theta=arctan(\frac{85}{100})=40.36\°$

Part 11)

Let

$\theta$ ----> angle of depression

we know that

$sin(\theta)=\frac{5,389-2,405}{3,044}$ ----> opposite side angle theta divided by hypotenuse

$sin(\theta)=\frac{2,984}{3,044}$

$\theta=arcsin(\frac{2,984}{3,044})=78.61\°$

Part 12) What is the exact value of arcsin(0.5)?

Remember that

$sin(30\°)=0.5$

therefore

$arcsin(0.5)$ -----> has two solutions

$arcsin(0.5)=30\°$ ----> I Quadrant

or

$arcsin(0.5)=180\°-30\°=150\°$ ----> II Quadrant

Part 13) What is the exact value of $arcsin(-\frac{\sqrt{2}}{2})$

The sine is negative

so

The angle lies in Quadrant III or Quadrant IV

Remember that

$sin(45\°)=\frac{\sqrt{2}}{2}$

therefore

$arcsin(-\frac{\sqrt{2}}{2})$ ----> has two solutions

$arcsin(-\frac{\sqrt{2}}{2})=-45\°$ ----> IV Quadrant

or

$arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\°$ ----> III Quadrant

5 0

3 years ago

20 At Honey Beepot, the bulk price for honey is $2.50 per pound, with a minimum purchase of 20 pounds. If Bobby paid $80 for som

Pachacha [2.7K]

Answer:
12 pounds (B)
Explanation:
80/2.5 = 32
32-20 = 12
I don’t know if this is right but I’m pretty sure it is.

7 0

3 years ago

Read 2 more answers

Nickolai lives at one end of Oregon Trail. Sydney lives at the other end of the trail. It is 5.8 kilometers from one end of Oreg

OleMash [197]

5.8 km - 2.79 km = 3.01 km

3.01 km = 3010 m

6 0

3 years ago

Read 2 more answers