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Free_Kalibri [48]

2 years ago

7

Find the positive square root of 217.05 and 100.2001......with division method

1 answer:

Alina [70]2 years ago

7 0

The square root of 217.05 is 14.73 and the square root of 100.2001 is 10.01.

<h3>How to depict the square root?</h3>

It should be noted that square root is factor of a number that when multiplied by itself will give the original number.

In this case, the square root of 217.05 is 14.73 and the square root of 100.2001 is 10.01.

Learn more about square root on:

brainly.com/question/428672

#SPJ1

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When the smaller of two consecutive integers is added to four times the larger, the result is 49. Find the integers.

ryzh [129]

$4(n+1)+n=49\\
4n+4+n=49\\
5n=45\\
n=9\\
n+1=10
$

These numbers are 9 and 10.

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3 years ago

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Answer:

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3 years ago

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(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

3 years ago

A pan of brownies is 7/10 full. Tyreese buys 2/5 of the brownies.

elixir [45]

Answer:

The answer to your question is below

Step-by-step explanation:

See the picture, please

In the first picture, we notice the initial amount of brownies (7/10), 7 brownies in a pan for 10 brownies.

After Tyreese bought 2/5 (7/10 - 2/5 = 3/10) there are 3 brownies left.

So in the picture, we notice only 3 brownies in a pan for 3 brownies.

6 0

3 years ago

Find the product.

Harrizon [31]

$(2n+2)(2n-2)\\\\=(2n)^2 -2^2 ~~~~~~;[a^2 -b^2 =(a-b)(a+b)]\\\\=4n^2 -4$

5 0

2 years ago

Read 2 more answers