A coin flipped 80 times . It lands tails 47 times . What is the P(heads)

Trina examined the equation below and said that the equation has exactly one solution. Negative 8 (x minus 4) + 2 x = negative 4

shutvik [7]

Answer

The answer is C. Trina is not correct because the two sides of the equation are equivalent.

Step-by-step explanation:

1. solve for x

-8(x - 4) + 2x = -4(x - 8) -2x

-8x + 32 + 2x = -4x + 32 -2x

-6x + 32 = - 6x + 32

0 = 0

This equation has infinite solutions, any rational number fits as solution for x in this equation. And as we can see both sides of it are equivalent.

Therefore, the correct answer is C. Trina is not correct because the two sides of the equation are equivalent.

Hope This Helps :p

8 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

What is the product? StartFraction x squared minus 16 Over 2 x + 8 EndFraction times StartFraction x cubed minus 2 x squared + x

zubka84 [21]

Answer:

[x(x - 1)(x - 4)]/(2(x + 4))

Step-by-step explanation:

We want to find;

[(x² - 16)/(2x + 8)] * [(x³ - 2x² + x)/(x² + 3x - 4)]

Now,

x² - 16 can be factorized as;

(x + 4)(x - 4)

Also, 2x + 8 can be factorized as;

2(x + 4)

Also, (x³ - 2x² + x) can factorized as;

x[x² - 2x + 1] = x[(x - 1)(x - 1)]

Also,(x² + 3x - 4) can be factorized out as; (x - 1)(x + 4)

So plugging in these factorized forms into the equation in the question, we have;

[(x + 4)(x - 4)/(2(x + 4))] * [x[(x - 1)(x - 1)] /((x - 1)(x + 4))

This gives;

((x - 4)/2) * x(x - 1)/(x +4)

This gives;

[x(x - 1)(x - 4)]/(2(x + 4))

8 0

2 years ago

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Bella's score on a spelling test is a function of the number of hours she studies. Which expression represents Bella's score on

Anna [14]

The correct answer is C

7 0

2 years ago

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Label each part of the equation. Write operation or variable N-100+r=54

Readme [11.4K]

The answer is n= 154 - r

6 0

2 years ago