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4 years ago

If one line has a slope of 2/3, then the slope of a line parallel to it would be?

Mathematics

1 answer:

ipn [44]4 years ago

8 0

-3/2

perpendicular slopes are those that are opposite sign and reciprocal of the original given slope

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A gift box in the shape of a rectangular prism has 20-inch length, 14-inch width, and 10-inch height. How much paper will you ne

Marizza181 [45]

Answer:

1240 in²

Step-by-step explanation:

Total surface area:

S = 2(lw + lh + wh)
S = 2(20*14+20*10+14*10) = 1240 in²

5 0

3 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

Answer number 3 please

alexgriva [62]

Answer:

10.50$

Step-by-step explanation:

if you take the 30 percent discount off of 35 dollars then you will get 10.50

3 0

3 years ago

How would I be able to answer this?

Ilya [14]

Radius = Diameter ÷ 2 = 10 ÷ 2 = 5mm

Find Volume: 

Volume= πr²h

Volume = π (5)² (6) = 471.24mm³ (Nearest hundredth)

Find Surface Area:

Surface area = 2πr² + 2πrh

Surface area = 2π(5)² + 2π(5)(6) = 345.58 mm²

Answer: Volume = 471.24mm³ , Surface area = 345.58 mm²

5 0

3 years ago

I need help factoring this expression

omeli [17]

X^2(x+2)-4(x+2)

(x^2-4) (x+2) = 0

do you have to keep going/?

or thats all they want?

3 0

3 years ago