If 6 identical articles can be bought for 2<br>17<br>Find the cost of each article.

SOMEONE PLEASE ANSWER AND EXPLAIN I’LL MARK BRAINLIEST

natita [175]

Answer:

Micheal=7x

Wife=2(7x)

Step-by-step explanation:

4 0

2 years ago

SOMEONE HELP PLEASE I WIL GIVE BRAinlest could yu please help answer 1-9 and I would really be grateful if you explained how you

ser-zykov [4K]

Answer: For the first one, 12.5% chance

Step-by-step explanation:

The probability of flipping a coin and getting tails is 50%. Then, if you were to flip two coins and get both tails, you would have to divide the 50% by 2, which is equals to 25% of flipping two coins and getting them both tails.

So now we know...

Flipping one coin gives you 50% tails and 25% of flipping two coins and getting two tails

So now, we do it again with the third coin. We would divide the 25% by 2, which will get us to a 12.5% of getting all tails, or in fractions, 1/8

I’m not exactly sure about the other ones but I hope you found this one helpful!

4 0

3 years ago

Read 2 more answers

Alphabet has how many letters?

LuckyWell [14K]

The alphabet has 26 letters

7 0

2 years ago

Read 2 more answers

The local home improvement store wants to increase their inventory. Last year 40 lawn mowers cost them $4,776.

hoa [83]

40 mowers = $4,776

Multiply the above equation by 3
120 mowers = $14,328

7 0

3 years ago

Read 2 more answers

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

If 6 identical articles can be bought for 217Find the cost of each article.​

If 6 identical articles can be bought for 217Find the cost of each article.