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2 years ago

Which graph represents the function p(x) = |x – 1|?

Mathematics

1 answer:

agasfer [191]2 years ago

7 0

"On a coordinate plane, an absolute value graph has a vertex at (1, 0)."

<h3>Which graph represents the absolute value function?</h3>

Remember that for a general absolute value function:

f(x) = |x - a| + b

The vertex is the point (a, b).

In this case we have:

p(x) = |x - 1|

Then the vertex is at the point (1, 0).

That is enough to conclude that the correct option is the last one:

"On a coordinate plane, an absolute value graph has a vertex at (1, 0)."

If you want to learn more about absolute values:

brainly.com/question/1782403

#SPJ1

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3 years ago

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8xq the product is between 0 and 8 when q equals 5/3 because 5/3 is greater than 1 and less than 2 true or false

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False. The answer should be between 8 and 16 because 5/3 = 1 2/3 which is between 1 & 2

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4 years ago

Using the quadratic formula to solve 4x2 – 3x + 9 = 2x + 1, what are the values of x? StartRoot 1 plus-or-minus StartRoot 159 En

castortr0y [4]

For this case we have that a quadratic equation is of the form:

$ax ^ 2 + bx + c = 0$

The roots are given by:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

We have the following equation:

$4x ^ 2-3x + 9 = 2x + 1\\4x ^ 2-3x-2x + 9-1 = 0\\4x ^ 2-5x + 8 = 0$

We look for the roots:

$x = \frac {- (- 5) \pm \sqrt {(- 5) ^ 2-4 (4) (8)}} {2 (4)}\\x = \frac {5 \pm \sqrt {25-128}} {8}\\x = \frac {5 \pm \sqrt {-103}} {8}$

We have to:

$i ^ 2 = -1$

So:

$x = \frac {5 \pm \sqrt {103i ^ 2}} {8}\\x = \frac {5 \pm i \sqrt {103}} {8}$

We have two imaginary roots:

$x_ {1} = \frac {5 \ + i \sqrt {103}} {8}\\x_ {2} = \frac {5 \ -i \sqrt {103}} {8}$

Answer:

$x_ {1} = \frac {5 \ + i \sqrt {103}} {8}\\x_ {2} = \frac {5 \ -i \sqrt {103}} {8}$

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4 years ago

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Mathematics MH helpo

Serjik [45]

<em><u>The inequalities that require flipping the sign are:</u></em>

-5x - 10 > 5

$\frac{x}{-7} + 3 \leq 4$

<em><u>Solution:</u></em>

Let us the inequalities one by one

You can perform on operations on both sides of inequality, and have its truth value unchanged

But if we multiply or divide by a negative number , we must flip the sign

<h3>option 1) </h3>

Move -10 from L.H.S to R.H.S

-5x > 5 + 10

-5x > 15

Divide the above expression by 5

$-x > 3$

Divide the above inequality by -1, so we must flip the sign

x < -3

<h3>option 2)</h3>

$7x - 5 \leq 16$

Move the constant term from L.H.S to R.H.S

$7x \leq 16 + 5\\\\7x \leq 21\\\\$

Divide the above inequality by 7

$x \leq 3$

This does not required flipping the symbol

<h3>option 3</h3>

$\frac{x}{5} - 6 > -11$

Move the constant term from L.H.S to R.H.S

$\frac{x}{5} > -11 + 6\\\\\frac{x}{5} > -5$

Multiply both the sides by 5

$x > -25$

This does not required flipping the symbol

<h3>option 4</h3>

$x + 12 \leq 29$

Move the constant term from L.H.S to R.H.S

$x \leq 29 - 12\\\\x \leq 17$

This does not required flipping the symbol

<h3>option 5</h3>

$\frac{x}{-7} + 3 \leq 4$

Move the constant term from L.H.S to R.H.S

$\frac{x}{-7} \leq 4-3\\\\\frac{x}{-7} \leq 1\\\\$

Multiply both the sides by -7, so we must flip the sign

$x \geq -7$

Thus this requires flipping the sign

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3 years ago

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