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3 years ago

–2 • 4? what is the anserw do you now

Mathematics

2 answers:

Contact [7]3 years ago

3 0

Answer:

Step-by-step explanation:

-8 because a negative times a positive is a negative so w times 4 equals I and since it’s a negative sign it’s -8

Alisiya [41]3 years ago

3 0

When u multiply by 1 negative your answer will be a negative so your answer will be -8

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3 years ago

A shirt originally cost $42.39, but it is an sale for $38.15. What is the percentage decrease of the price of the shirt? If nece

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3 years ago

Solve the inequality.

marysya [2.9K]

Answer:

$\huge x > \frac{7}{3} \\$

Step-by-step explanation:

$8( \frac{1}{2} x - \frac{1}{4} ) > 12 - 2x \\$

Expand the terms in the bracket

That's

$8( \frac{1}{2} x) - 8( \frac{1}{4} ) > 12 - 2x \\ 4x - 2 > 12 - 2x$

Move -2x to the other side of the inequality

$4x + 2x - 2 > 12 \\ 6x - 2 > 12$

Move - 2 to the other side of the inequality

$6x > 12 + 2 \\ 6x > 14$

Divide both sides by 6

$\frac{6x}{6} > \frac{14}{6} \\$

We have the final answer as

$x > \frac{7}{3} \\$

Hope this helps you

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3 years ago

Identify the interval on which the quadratic function is positive.

Alenkasestr [34]

Answer:

$\textsf{1. \quad Solution: $1 < x < 4$,\quad Interval notation: $(1, 4)$}$

$\textsf{2. \quad Solution: $-2 < x < 4$,\quad Interval notation: $(-2, 4)$}$

Step-by-step explanation:

<h3>Question 1</h3>

The intervals on which a quadratic function is positive are those intervals where the function is above the x-axis, i.e. where y > 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a negative leading coefficient, the parabola opens downwards. Therefore, the interval on which y > 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies -7x^2+35x-28& = 0\\-7(x^2-5x+4)& = 0\\x^2-5x+4& = 0\\x^2-x-4x+4& = 0\\x(x-1)-4(x-1)&= 0\\(x-1)(x-4)& = 0\end{aligned}$

Apply the zero-product property:

$\implies x-1=0 \implies x=1$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is positive is:

Solution: 1 < x < 4
Interval notation: (1, 4)

<h3>Question 2</h3>

The intervals on which a quadratic function is negative are those intervals where the function is below the x-axis, i.e. where y < 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a positive leading coefficient, the parabola opens upwards. Therefore, the interval on which y < 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies 2x^2-4x-16& = 0\\2(x^2-2x-8)& = 0\\x^2-2x-8& = 0\\x^2-4x+2-8& = 0\\x(x-4)+2(x-4)&= 0\\(x+2)(x-4)& = 0\end{aligned}$

Apply the zero-product property:

$\implies x+2=0 \implies x=-2$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is negative is:

Solution: -2 < x < 4
Interval notation: (-2, 4)

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