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3 years ago

Which of the following functions is represented by the graph below?

Mathematics

1 answer:

Ann [662]3 years ago

6 0

Answer:

A. y = 3 cos (1/2) x.

Step-by-step explanation:

The y-intercept is 3 so it will contain 3 cos x (because y = cos x passes through (0, 1).

The graph of 3 cos x has been stretched horizontally by a factor 2 ( 3 cos x would pass through (π/2, 0) whereas the given graph passes through (π, 0).

So the equation is 3 cos(1/2) x

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What is the order of operation for 2(9 + 7)?

shepuryov [24]

the answer is 32 because you add 9 plus 7 then you multiply tha th # by 2

6 0

3 years ago

Read 2 more answers

You are choosing between two health clubs. Club A offers membership for a fee of $ 28 plus a monthly fee of $ 22. Club B offers

Mazyrski [523]

Answer: 2 months, Total cost= $72

Step-by-step explanation:

Let x denote the number of months then

For Club A fee= 28+22x

For Club B fee= 16+28x

Equating both we get

28 + 22x= 16+28x

28-16= 28-22x

12= 6x

x= 2

So after 2 months the fee will be same i.e. 28+22(2)= $72

8 0

3 years ago

1) The deer population in a national park is expected to decline over time. The park

Lorico [155]

Answer:

Here's what I find.

Step-by-step explanation:

You have 800 deer at the end of Year 1, and you expect the population to decrease each year thereafter.

a) i) The recursive formula

Let dₙ = the deer population n years after the initial measurement.

$d_{n} = d_{n - 1}r^{n}$

For this situation,

$d_{n} = d_{n - 1}(0.5)^{n}$

a) ii) Definitions

n = the number of years from first measurement

r = the common ratio, that is, the deer population at the end of one year divided by the population of the previous year.

a) iii) First term of sequence

The first term of the sequence is d₀, the population when first measured.

b) The function formula

The formula for the nth term of a geometric series is

$d_{n} =d_{0}r^{n - 1}$

c) Value of d₀

Let n = 2; then d₂ = 800

$\begin{array}{rcl}800 & = & d_{0}(0.5)^{2 - 1}\\800 & = & d_{0}(0.5)\\\\d_{0} & = & \dfrac{800}{0.5}\\\\& =&\mathbf{1600}\\\end{array}$

4 0

3 years ago

Help me with differentation and integration please!!

Marina86 [1]

Answer:

See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

$\dfrac{d}{dx}\tan x=\sec^2$

Using the chain rule

$\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}$

such that $u = \tan x$

we can get a general formulation for

$y = \tan^n x$

Considering the power rule

$\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$

we have

$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

therefore,

$\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x$

Now, once

$\sec^2 x - 1= \tan^2x$

we have

$3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x$

Hence, we showed

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0

3 years ago

QUESTION: Use the two way table below to answer the question: How

qaws [65]

Answer:

B.91

Step-by-step explanation:

Because it says so on the table.

7 0

3 years ago