Layla used 0.482 gram of salt in her experiment. Maurice use 0.51 gram of salt. who used the greater amount of salt?

A dust particle weighs 7.42 × 10-10 kilograms. What is the weight of 5 × 106 dust particles represented in scientific notation?

vova2212 [387]

5 x 106 = 530 particles

530 x 7.42x10^-10 = 3.93x10^-7

7 0

2 years ago

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Giuseppe is studying shark attacks around the world and finds that most such attacks occur during the day. He concludes that sha

HACTEHA [7]

Answer:

C) not justified, since people usually swim in the ocean during the day.

5 0

2 years ago

What is the range of the exponential function shown below

Vsevolod [243]

Answer : y>0

f(x) = 9*2^x

f(x) is an exponential function

$f(x) = 9*2^x$

When we plug in positive value for x , the value of y is positive

When we plug in negative value for x , the value y is also positive

So for any value of x, the y value is positive always.

Range is the set of y values for which the function is defined

y values are positive , so range is y >0

3 0

2 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

2 years ago

Compare A and B if 70% of A is equal to $0.7a and 35% of B is equal to $0.35a

MrRa [10]

Answer:

A=B

Step-by-step explanation:

Equation 1: 70% of A = 0.7a

Equation 2: 35% of B = 0.35a

Equation 3 = (Equation 2) • 2

Equation 3: 70% of B = 0.7a

Equation 3 = Equation 1

A = B

3 0

2 years ago

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