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3 years ago

Why the product (-6)(-3) is positive

Mathematics

1 answer:

rewona [7]3 years ago

7 0

Answer:

hope this helps

have a good day :)

Step-by-step explanation:

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A rectangular prism has a volume of 16 cm3 and a total surface area of 40 cm2. A similar prism has length 10 in., width 5 in., a

Kruka [31]

We know that
[volume of the similar prism]=10*5*5----> 250 in³
volume original prism=16 cm³
1 in³---------> 16.3871 cm³
X----------> 16
x=16/16.3871------> x=0.9764 in³

[volume of the similar prism]=[scale factor]³*[volume original prism]
[scale factor]³=[volume of the similar prism]/[volume original prism]
[scale factor]³=[250]/[0.9764]------> 256.05
scale factor=∛256.05-------> 6.35

the dimensions of the original prism are
1 in-----> 2.54 cm

length 10 in/6.35-------> 1.57 in*2.54 cm/in-----> 4 cm
width 5 in/6.35---------> 0.79 in*2.54 cm/in-----> 2 cm
height 5 in/6.35--------> 0.79 in*2.54 cm/in----> 2 cm

3 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

Help wit this? 10m-8m+2+10

GenaCL600 [577]

Answer:

2m+12

Step-by-step explanation:

add like terms

2m+12

7 0

3 years ago

(-2) times (+5) times (-1) times (-5) times (-2) please can u answer fast i have a test in 2 min

zlopas [31]

Answer:

100

Step-by-step explanation:

4 0

10 months ago

The solutions to the inequality ys-x+1 are shaded on

kupik [55]

Answer:

the solutions to the inequality ys-x+1 are shaded on the graph. which point is B. (3 ,-2)

6 0

3 years ago