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2 years ago

B One of these numbers is the area of the outside curved surface of the cylinder, in cm', using

Mathematics

1 answer:

GrogVix [38]2 years ago

6 0

Answer:

$area = 2\pi \: rh \\ = 2 \times 3 \times 5 \times 1 \\ = 30 \: {cm}^{2}$

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Find the coordinates of k

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If you plot the coordinates on a piece of graph paper your answer would come out too (1,-8)

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3 years ago

What is the answer and how did you work it out

Anna71 [15]

Answer: b

Step-by-step explanation:

10 x 10= 100

15 x 10= 150

100+150= 250

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3 years ago

A segment can have more than one bisector. True False

insens350 [35]

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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:

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