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3 years ago

What is the average of the integers from 25-41

2 answers:

7 0

The average of 25 and 41 is (1/2)(25+41) = (1/2)(66) = 33 .

Similarly ...
--The average of 26 and 40 is 33.
-- The average of 27 and 39 is 33.
-- The average of 28 and 38 is 33.
-- The average of 29 and 37 is 33.
-- The average of 30 and 36 is 33.
-- The average of 31 and 35 is 33.
-- The average of 32 and 34 is 33.
-- That leaves only 33.

So the average of all of those 17 integers is 33 .

Mashcka [7]3 years ago

4 0

The answer is 33 for your question

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Which expression is equivalent to (x Superscript 27 Baseline y) Superscript one-third?.

marin [14]

To solve the problem we must know the basic exponential properties.

<h3>What are the basic exponent properties?</h3>

${a^m} \cdot {a^n} = a^{(m+n)}$

$\dfrac{a^m}{a^n} = a^{(m-n)}$

$\sqrt[m]{a^n} = a^{\frac{n}{m}}$

$(a^m)^n = a^{m\times n}$

$(m\times n)^a = m^a\times n^a$

The expression can be written as $x^9\sqrt[3]{y}$ .

Given to us

$(x^{27}y)^\frac{1}{3}$

$(x^{27}y)^\frac{1}{3}$

Using the exponential property $(m\times n)^a = m^a\times n^a$ ,

$=(x^{27}y)^\frac{1}{3}\\\\=x^{\frac{27}{3}}\times y^\frac{1}{3}\\\\=x^9\times y^\frac{1}{3}$

Using the exponential property $\sqrt[m]{a^n} = a^{\frac{n}{m}}$ ,

$=x^9\times y^\frac{1}{3}\\\\=x^9\times \sqrt[3]{y}\\\\=x^9 \sqrt[3]{y}$

Hence, the expression can be written as $x^9\sqrt[3]{y}$ .

Learn more about Exponent properties:

brainly.com/question/1807508

5 0

2 years ago

Read 2 more answers

(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

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

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

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

$=5x^4y$

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

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

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

Can someone please help

Sonja [21]

Answer:

i have no idea lol make it more clearer

6 0

3 years ago

In mathematics, the distance between one point (a) and another point (b), each with coordinates (x,y), can be computed by taking

anygoal [31]

The formula to find the distance between points $P_{1}$ and $x_{2}$ is given as $\sqrt{ ( y_{1}- y_{2}) ^{2}+ ( x_{1}- x_{2}) ^{2} }$ , where

$y_{1}- y_{2}$ is the vertical distance between two points on the y-axis
$x_{1} - x_{2}$ is the horizontal distance between two points on the x-axis

8 0

3 years ago

I really need the help.....

prisoha [69]

Answer:

Similar - Top center, bottom left

Not similar, different shape - Top left, bottom right

Not similar, different ratio of side lengths - Top right, bottom center

Step-by-step explanation:

We can see automatically that the top left and bottom right are not similar because they are a triangle and a trapezoid, not a rectangle.

Top right is not similar because of its different ratio of sides. The ratio for the sides of the original rectangle is 3:7, while the ratio of the top right is 2:3.5 .

Bottom center is not similar because of its different ratio of sides as well, as its ratio is 6:12

6 0

3 years ago