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3 years ago

If Julia has 20 cupcakes, and 10 people eat one cupcake, how much cupcakes are left?

Mathematics

2 answers:

Feliz [49]3 years ago

8 0

There are going to be 10 cupcakes left,hope I helped (:

Mila [183]3 years ago

3 0

There will be 10 cupcakes left hope that helped see ya

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Maria took home a net pay of $1,200 on her paycheck this week. She worked 60 hours during this specific pay period and had deduc

erastovalidia [21]

Answer:

$25 per hour

Step-by-step explanation:

Net pay= $1200

Deductions= $300

Hours worked= 60

Total pay=$1200+ $300= $1500.

Since her net pay is the money left after deductions, so we add the netpay and deductions to get her total pay.

Therefore, to find her hourly pay, we simply divide the total pay by the hours worked,so we have 1500/60= $25/hour

8 0

3 years ago

Read 2 more answers

A rectangular floor covered without overlaps or gaps by 7 identical tiles, each of the length l and with w. If the w=12 ft, what

SOVA2 [1]

A rectangular floor covered without overlaps or gaps by 7 identical tiles, each of the length l and with w. If the w=12 ft, what is the total area of the floor?

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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,

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

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

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3 years ago

What is 120÷37 please show the work and answer

Firlakuza [10]

$120/37$

$=3.2$

$or =3.0$

Hope this helps!

Thanks!

-Charlie

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3 years ago

Explain how you determine whether your estimate is an overestimate or underestimate

Montano1993 [528]

If your number is smaller than the actual number, you underestimated. If it is larger, you overestimated. Please mark Brainliest!!!

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4 years ago