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3 years ago

The graph below shows distance to time for one of Anna's swimming races. What is Anna's average speed.

Mathematics

1 answer:

DedPeter [7]3 years ago

7 0

Answer:

Step-by-step explanation:

In this case, on each coordinate point, distance/time = 2.

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A workers regular hourly rate of pay is $13. What is the workers total pay for working 40 regulars and 3 hours overtime at a rat

ANTONII [103]

Answer:

578.50

Step-by-step explanation:

The worker worked 40 hours at 13 dollars

40 * 13 =520

Then they worked 3 hours at 13(1.5) or 19.50

3*19.50 =58.5

The total amount of money earned is regular time plus overtime

520 + 58.5 =578.50

3 0

2 years ago

Again the auction house raffle sold 200 tickets at $1 each. How many prizes of $10 should they to make the raffle a fair game?

jekas [21]

200; $1e\pp

they should sell 20 prizes

20x10=200

5 0

2 years ago

Read 2 more answers

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

2 years ago

HELP ASAP

almond37 [142]

Answer:

Step-by-step explanation:

let the number of dimes=x

and number of quarters=y

x=2y

0.10 x+0.25 y=2.25

or 10x+25y=225

divide by 5

2x+5y=45

2 ×2y+5y=45

4y+5y=45

9y=45

y=45/9=5

x=2y=2×5=10

Hence dimes=10

quarters=5

6 0

3 years ago

Find b: a2+b2=c2<br> a = 6, c = 10<br><br> PLZ HELP YALL ILL REWARD BRAINLIEST THINGY

Kaylis [27]

Answer:

b = 4

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers