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3 years ago

How many 1/3 yard lengths are in 1 yard

Mathematics

2 answers:

Jobisdone [24]3 years ago

8 0

Answer:

Step-by-step explanation:

There are three 1/3 yard lengths in 1 yard since we cut it into three pieces.

Wewaii [24]3 years ago

6 0

Answer:

Step-by-step explanation:

1/3 yard lengths times 3 equals 3 (or 1 yard).

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Ms Lucas has 9 cases of crayon with 52 boxes in each case Ms Bruns has 6 cases of crayon with a 104 in each case How many boxes

saveliy_v [14]

Answer:

1092

Step-by-step explanation:

Step one:

given data

Ms Lucas has 9 cases of crayon with 52 boxes in each case

total crayon is

9*52= 468

Ms Bruns has 6 cases of crayon with a 104 in each case

total crayon is

6*104= 624

The total crayon is

468+624= 1092

5 0

3 years ago

An arrangement of flowers originally priced at $48.00 is marked down to $39.00. What is the percent of decrease, rounded to the

Mashcka [7]

The answer is C

So how?

I multiplied 48*.19= 9.12 then subtracted that from 48 and got 38.88 which rounds to 39

C is the answer

I hope this helps! :D

6 0

3 years ago

Billy joined Franco's today. His membership cost him $50. He has to pay $2 each time he goes.

Darina [25.2K]

Answer:

Step-by-step explanation:

M is the membership and b is billy

3 0

3 years ago

73.4 round each number to the place of the underlined digit underline number is 7

marishachu [46]

Answer:

70.

Step-by-step explanation:

4 0

2 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

2 years ago