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3 years ago

Which graph represents the system of inequalities? y+2<5xy+3x>4y≥2x

Mathematics

2 answers:

Lelechka [254]3 years ago

8 0

The graph that represents the system of inequalities is graph 3

IgorC [24]3 years ago

5 0

Answer:

The correct option is 2.

Step-by-step explanation:

The given system of inequalities

$y+2$

$y+3x>4$

$y\geq 2x$

Inequality (1) can be written as

$y$

The related equation is

$y=5x-2$

The slope of line is 5 and y-intercept is -2.

The sign of inequality is < , it means related line of inequality (1) is a dotted line and shaded region is below the line.

Inequality (2) can be written as

$y>-3x+4$

The related equation is

$y=-3x+4$

The slope of line is -3 and y-intercept is 4.

The sign of inequality is >, it means related line is a dotted line and shaded region is above the line.

Inequality (3) is

$y\geq 2x$

The related equation is

$y=2x$

The slope of line is 2 and y-intercept is 0.

The sign of inequality is ≥, it means related line is a solid line and shaded region is above the line.

In the below graph the common shaded region is the solution region.

Therefore the correct option is 2.

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2 years ago

Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.

loris [4]

To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>

$x^ax^b = x^{(a+b)}$
$\dfrac{x^a}{x^b} = x^{(a-b)}$
$(a^a)^b =x^{(a\times b)}$
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<h2>Explanation</h2>

Given to us

$(16x^8y^{12})^{\frac{1}{2}}$

Solution

We know that 16 can be reduced to $2^4$ ,

$=(2^4x^8y^{12})^{\frac{1}{2}}$

Using identity $(xy)^a = x^ay^a$ ,

$=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}$

Using identity $(a^a)^b =x^{(a\times b)}$ ,

$=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})$

Solving further

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Learn more about Exponentiation:

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2 years ago

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Answer: Could you check if you can make the question more visible?

Step-by-step explanation:

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3 years ago

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uranmaximum [27]

Hey there!

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I hope this helps!

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3 years ago

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