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2 years ago

8

Bill went on a very long run, so he ran kind of slowly. He ran 2 miles in 30 minutes. How many miles would he run in three hours

?

1 answer:

FromTheMoon [43]2 years ago

7 0

Answer:

12 miles

Step-by-step explanation:

There are 3 hours = 180 minutes

180÷30=6

6×2=12

Hope that helps!

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2 years ago

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3 years ago

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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><u>Question 1</u></h3>

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

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

As the given <u>quadratic function</u> 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 <u>quadratic function</u>, 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 <u>zero-product property</u>:

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

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

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

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

<h3><u>Question 2</u></h3>

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

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

As the given <u>quadratic function</u> 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 <u>quadratic function</u>, 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 <u>zero-product property</u>:

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

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

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

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

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1 year ago