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3 years ago

13

If you apply ghe change below to the quadratic parent function f(x)=x2 what is the equation of the new function? Shift 1 unit ri

ght vertically stretch by a factor of 5 reflect over the x-axis

1 answer:

saveliy_v [14]3 years ago

6 0

1 to left gives (x -1)^2

vertical stretch makes it 5(x - 1)^2

reflection in axis changes the sign to - :-

equation of new function is f(x) = -5(x - 1)^2

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Is .634 rational number

Ksivusya [100]

Yes 0.634 is a rational number.
Rational numbers are those numbers that can be still expressed in standard form or in fraction form and vice-versa. Unlike irrational numbers that are opposed to the definition of rational numbers. These values include pi, square root of two and etc. These values are impossible to fractionize. To better illustrate this circumstance.

We can have calculate a number that will have a quotient of 0.634 or a fraction that is equal to the given value. <span><span>
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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 $\frac{\sqrt[3]{16y^4}}{x^2}$

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3 years ago

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Float

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3 years ago

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Which set of ordered pairs represents y as a function of x?<br> PLEASE HELP 15 POINTS OR BRAINLIEST

gizmo_the_mogwai [7]

Answer:

A

Step-by-step explanation:

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2 years ago

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What is the answer to this equation t+t+t=12

OLEGan [10]

Answer:

t = 4

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS
Equality Properties

<u>Algebra I</u>

Combining Like Terms

Step-by-step explanation:

<u>Step 1: Define equation</u>

t + t + t = 12

<u>Step 2: Solve for </u><em><u>t</u></em>

Combine like terms (t): 3t = 12
Divide 3 on both sides: t = 4

<u>Step 3: Check</u>

<em>Plug in t into the original equation to verify it's a solution.</em>

Substitute in <em>t</em>: 4 + 4 + 4 = 12
Add: 8 + 4 = 12
Add: 12 = 12

Here we see that 12 does indeed equal 12.

∴ t = 4 is a solution of the equation.

3 0

3 years ago