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3 years ago

Neither the absolute value function nor the greatest integer function is one-to-one.

1 answer:

4 0

It is true that neither the absolute value function nor the greatest integer function is one to one. The correct answer is True.

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What is the following product?

devlian [24]

Answer: Last Option

$4x^5\sqrt[3]{3x}$

Step-by-step explanation:

To make the product of these expressions you must use the property of multiplication of roots:

$\sqrt[n]{x^m}*\sqrt[n]{x^b} = \sqrt[n]{x^{m+b}}$

we also know that:

$\sqrt[3]{x^3} = x$

$\sqrt[3]{16x^7}*\sqrt[3]{12x^9}\\\\\sqrt[3]{16x^3x^3x}*\sqrt[3]{12(x^3)^3}\\\\x^2\sqrt[3]{16x}*x^3\sqrt[3]{12}\\\\x^5\sqrt[3]{16x*12}\\\\x^5\sqrt[3]{2^4x*2^2*3}\\\\x^5\sqrt[3]{2^6x*3}\\\\4x^5\sqrt[3]{3x}$

6 0

4 years ago

Determine whether y = 2x - 3 is a linear equation. If so, write the equation in standard form.

beks73 [17]

Answer:

yes, y=2x-3 is linear

2x - y = 3 Standard Form

Step-by-step explanation:

since both the 'x' and 'y' variables have one as their exponents then this represents a linear equation

standard form: Ax + By = C

y = 2x - 3

subtract 2x from each side to get:

-2x + y = -3

divide each term by -1 to get (this is done so the leading coefficient is not negative)

2x - y = 3 Standard Form

4 0

3 years ago

Read 2 more answers

Which is a perfect square?

velikii [3]

Answer:

6 squared

Step-by-step explanation:

The square root of a perfect square is a rational number.

4 0

3 years ago

A ball is thrown vertically upward from the ground with an initial velocity of 111 ft/sec. Use the quadratic function h(t) = −16

ddd [48]

Answer:

t=3.5 seconds

Step-by-step explanation:

Given

h(t) = −16t^2 + 111t + 0

h'(t)= -32t + 111

Maximum height occurs when h(t) = 0 and the ball begins to fall

h(t)= -32t + 111=0

-32t + 111=0

-32t=-111

Divide both sides by -32

t=3.46872

Approximately, t=3.5 seconds

Recall,

Maximum height occurs when h(t) = 0

h(t)= -32t + 111=0

= -32(3.46872)+111

= -110.99904+111

= 0.00096 ft

4 0

3 years ago

Long Division: (3x^4-8x^3-x^2+9x+5) / (x-2) pleaseee answer this

Drupady [299]

3x⁴ = 3x³ • x, and

3x³ (x - 2) = 3x⁴ - 6x³

Subtract this from the numerator to get a remainder of

(3x⁴ - 8x³ - x² + 9x + 5) - (3x⁴ - 6x³) = -2x³ - x² + 9x + 5

-2x³ = -2x² • x, and

-2x² (x - 2) = -2x³ + 4x²

Subtract this from the previous remainder to get a new remainder of

(-2x³ - x² + 9x + 5) - (-2x³ + 4x²) = -5x² + 9x + 5

-5x² = -5x • x, and

-5x (x - 2) = -5x² + 10x

Subtract this from the last remainder to get a new one of

(-5x² + 9x + 5) - (-5x² + 10x) = -x + 5

-x = -1 • x, and

-1 (x - 2) = -x + 2

This gives a new remainder of

(-x + 5) - (-x + 2) = 3

3 does not divide x, so we're done.

So, we have

(3x⁴ - 8x³ - x² + 9x + 5) / (x - 2) = 3x³ - 2x² - 5x - 1 + 3/(x - 2)

7 0

3 years ago