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2 years ago

A system of equations has infinitely many solutions. If 2y – 4x = 6 is one of the equations, which could be the other equation?

Mathematics

1 answer:

Molodets [167]2 years ago

5 0

Hi there!

$\large\boxed{y = 2x + 6}$

For a system to have infinite solutions, the lines have to be the same. We can begin by rearranging the given equation into the format y = mx + b:

2y - 4x = 6

Move x to the opposite side:

2y = 4x + 6

Divide all terms by 2:

y = 2x + 6

The answer choice that matches this is the first one.

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Is point B a reflection across the x-axis of point C? Explain why or why not.

Black_prince [1.1K]

Answer:

no ot is not b/c both would be negative and its not the same points as B

5 0

1 year ago

Y=(1/2)^x +2. What is the horizontal asymptote of this function?

professor190 [17]

$\lim_{x\to -\infty} \left(\frac{1}{2}\right)^x+2=\infty+2=\infty\\
\lim_{x\to \infty} \left(\frac{1}{2}\right)^x+2=0+2=2\\$

There is one-sided horizontal asymptote $y=2$

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3 years ago

5.804 rounded to the nearest hundred

nalin [4]

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7 0

2 years ago

Read 2 more answers

Point K is on line segment JL. Given KL = 2x – 2, JL = 4x + 9, and

sergeinik [125]

Answer:

JL = 21

Step-by-step explanation:

Given that K is on line segment JL, therefore:

KL + JK = JL (according to segment addition postulate)

KL = 2x - 2

JK = 5x + 2

JL = 4x + 9

Thus:

$(2x - 2) + (5x + 2) = (4x + 9)$

Solve for x

$2x - 2 + 5x + 2 = 4x + 9$

$2x +5x - 2 + 2 = 4x + 9$

$7x = 4x + 9$

Subtract 4x from both sides

$7x - 4x = 4x + 9 - 4x$

$3x = 9$

Divide both sides by 3

$\frac{3x}{3} = \frac{9}{3}$

$x = 3$

Find the numerical length of JL

$JL = 4x + 9$

Plug in the value of x

$JL = 4(3) + 9 = 12 + 9 = 21$

7 0

2 years ago

Find f'(x) and state the domain of f': f(x) = In (2x^2+1)

-Dominant- [34]

Answer:

$f'(x) = \frac{4x}{2x^2+1}$

Domain: All Real Numbers

General Formulas and Concepts:

Algebra I

Domain is the set of x-values that can be inputted into function f(x)

Calculus

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Derivative: $\frac{d}{dx} [ln(u)] = \frac{u'}{u}$

Step-by-step explanation:

Step 1: Define

f(x) = ln(2x² + 1)

Step 2: Differentiate

Derivative ln(u) [Chain Rule/Basic Power]: $f'(x) = \frac{1}{2x^2+1} \cdot 2 \cdot 2x^{2-1}$
Simplify: $f'(x) = \frac{1}{2x^2+1} \cdot 4x$
Multiply: $f'(x) = \frac{4x}{2x^2+1}$

Step 3: Domain

We know that we would have issues in the denominator when we have a rational expression. However, we can see that the denominator would never equal 0.

Therefore, our domain would be all real numbers.

We can also graph the differential function to analyze the domain.

5 0

2 years ago