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AlladinOne [14]

3 years ago

8

If the first and third of three consecutive even integers are added, the result is 22 less than three times the second integer.

Find the integers

1 answer:

icang [17]3 years ago

4 0

Three consecutive even integers are 20, 22 and 24

<em><u>Explanation</u></em>

Suppose, the three consecutive even integers are $n, n+2$ and $n+4$

Given that, sum of first and third integers is 22 less than 3 times the second integer. So, the equation will be....

$n+(n+4)= 3(n+2)-22\\ \\ 2n+4=3n+6-22\\ \\ 2n-3n=6-22-4\\ \\ -n=-20 \\ \\ n=20$

So, the first integer is 20, the second integer is (20+2)= 22 and the third integer is (20+4)= 24

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Brilliant_brown [7]

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6 0

3 years ago

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which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

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

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

Simplify and show step by step 30 POINTS!

PIT_PIT [208]

Step-by-step explanation:

the explanation is in the diagram above

4 0

2 years ago

Kelly needs $125 to buy a dress for the dance. Her dad gave her five dollars for doing your chores and received $15 an hour to t

Debora [2.8K]

8 hours. 125 - 5 = 120. 120 / 15 = 8. Kelly will need to tutor for 8 hours minimum to buy the dress.

7 0

3 years ago

Which is the graph of the following equation?

lesantik [10]

Given equation is

$\frac{(x-5)^2}{16} - \frac{(y-2)^2}{9}=1$

The given equation is in the form of

$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}=1$

a^2= 16 , so a=4

b^2 = 9 so b= 4

The value of 'a' is greater than the value of 'b'

So it is a Horizontal hyperbola

First two graphs are horizontal hyperbola

Here center is (h,k)

h= 5 and k =2 from the given equation

So center is (5,2)

Now we find vertices

Vertices are (h+a,k) and (h-a,k)

We know h=5, k=2 and a=4

So vertices are (9,2) and (1,2)

Second graph having same vertices and center

The correct graph is attached below

7 0

3 years ago

Read 2 more answers