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viva [34]
3 years ago
6

What are some ways you can make a rectangle 10,000 cm squared (I NEED A AWNSER QUICK)​

Mathematics
1 answer:
lidiya [134]3 years ago
8 0

Answer:

2 longer sides would be 4000 each, so that's 8000 total

and the 2 shorter sides could be 1000 each, so that's 2000 total

8000+2000=10,000

Step-by-step explanation:

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Subtract −7n+2 ​​from 2n−1 ​.
Readme [11.4K]

Answer:

Subtracting  −7n+2 ​​from 2n−1 ​ is 9n - 3.

Step-by-step explanation:

As the expression given in the question be as follow.

−7n+2 and 2n - 1 .

Subtracted −7n+2 from 2n - 1 .

= 2n - 1 - (-7n + 2)

Open the bracket

= 2n - 1 + 7n - 2

Simplify the above

= 9n - 3

Therefore Subtracting  −7n+2 ​​from 2n−1 ​ is 9n - 3.

6 0
3 years ago
Read 2 more answers
Mr Day has 113 frog stickers. He wants to give 10 frog stickers to each student. How many students will get 10 stickers? will th
Kazeer [188]
11 student will get 10  stickers and 3 will be <span>left over.</span>
5 0
3 years ago
Read 2 more answers
“encontrar la integral indefinida y verificar el resultado mediante derivación”
Oliga [24]

I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx

Haz la sustitución:

y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx

\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C

Para confirmar el resultado:

\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}

I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx

Sustituye:

y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx

\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C

(Te dejaré confirmar por ti mismo.)

I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx

Sustituye:

y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx

\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C

I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}

Sustituye:

u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}

\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C

Podemos hacer que esto se vea un poco mejor:

\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}

\implies I=-\dfrac{(t+1)^4}{4t^4}+C

4 0
3 years ago
Subtract. 0.63–(–9.85)<br> a. –10.48<br> b. –9.22<br> c. 9.22<br> d. 10.48
zaharov [31]
Negative times negative = positive.

=0.63−(−9.85)
=0.63+9.85
=10.48

D) is the answer.
5 0
3 years ago
Read 2 more answers
PLEASE PLEASE HELP ME
mina [271]
The answer is <EKF and <HKI
3 0
3 years ago
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