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3 years ago

A grid shows the position of a subway stop at your house. The subway stop is located at (-5, 2) and your house is located (-9,9)

. What is the distance, to the nearest unit, between your house and the subway stop?

Mathematics

1 answer:

aleksley [76]3 years ago

3 0

$A(-5;\ 2);\ B(-9;\ 9)\\\\|AB|=\sqrt{(-9-(-5))^2+(9-2)^2}=\sqrt{(-4)^2+7^2}=\sqrt{16+49}\\\\=\sqrt{65}\approx8\ (units)$

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While bowling with friends, Brandy rolls a strike in 57 out of 72 frames. what is the experimental probability that brandy will

gogolik [260]

Answer:

79.17% probability that brandy will roll a strike the first frame of the next game

Step-by-step explanation:

The experimental probability is the probability based on previous outcomes, that is, the number of desired outcomes divided by the number of total outcomes.

While bowling with friends, Brandy rolls a strike in 57 out of 72 frames. What is the experimental probability that brandy will roll a strike the first frame of the next game?

Desired outcomes:

57 strikes

Total outcomes:

72 frames

Probability:

p = 57/72 = 0.7917

79.17% probability that brandy will roll a strike the first frame of the next game

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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:

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