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3 years ago

Multiply. 6 1/4 x 3 3/5 =

Mathematics

1 answer:

Shalnov [3]3 years ago

6 0

45/2 or 22 1/2
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Please solve for y X=2y-3

inessss [21]

Answer:

y = $\frac{x+3}{2}$

Step-by-step explanation:

Given

x = 2y - 3 ( add 3 to both sides )

x + 3 = 2y ( divide both sides by 2 )

$\frac{x+3}{2}$ = y

6 0

4 years ago

Variable p is 2 more than variable d. Variable p is also 1 less than variable d. Which pair of equations best models the relatio

romanna [79]

MY answer is p= d + 1d = 2p

3 0

4 years ago

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

Evaluate: 12x+10 for x=−12

julsineya [31]

Answer:

-144?

Step-by-step explanation:

im not sure if that'll be the answer

5 0

3 years ago

Read 2 more answers

For what side length(s) is the area of an equilateral triangle equal to 30 cm?? Only enter the number, in centimeters, rounded t

xenn [34]

Answer: The sides length are 8.32 cm

Step-by-step explanation:

An equilateral triangle has all his sides of the same lenght, so we assume that the triangle has an L lenght in his sides.

The area of a triangle is $Area = \frac{base * height}{2}$ where the base is L, the Area is 30 and an unknown height.

To determine the height, we cut the triangle in half and take one side. By simetry, one side has a base of $\frac{L}{2}$ , a hypotenuse of L and a the unknown height.

Then we apply the Pythagoras theorem, this states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides, or, $hypotenuse = \sqrt{c^{2} + c^{2} }$ Where one c is $\frac{L}{2}$ and the other is the height.

Then we find one of the c of the equation wich will be the height.

$height = \sqrt{hypotenuse^{2}-base^{2} }$

$height = \sqrt{ L^{2} -\frac{L}{4} ^{2}}\\height = \sqrt{\frac{ 3L^{2}}{4} } \\\\height = \frac{\sqrt{3}L }{2}$

Finally, we use the triangle area mentioned before an find the value of L.

$30 = \frac{L*\frac{\sqrt{3}L }{2} }{2} \\\\L = \sqrt{\frac{120}{\sqrt{3} } } \\\\L = 8.32 cm$

6 0

3 years ago