Can someone please simplify these using product notation?

What is the chemical formula for the molecule modeled?

adelina 88 [10]

The correct answer is A

8 0

2 years ago

There are 24 more students in the seventh grade class than the number g in the eighth grade class. The seventh grade class has 1

Over [174]

Answer:

160-24=g

Step-by-step explanation:

6 0

2 years ago

What are 3 completed examples for functions?

Lyrx [107]

Example 1

Write y = x2 + 4x + 1 using function notation and evaluate the function at x = 3.

Solution

Given, y = x2 + 4x + 1

By applying function notation, we get

f(x) = x2 + 4x + 1

Evaluation:

Substitute x with 3

f (3) = 32 + 4 × 3 + 1 = 9 + 12 + 1 = 22

Example 2

Evaluate the function f(x) = 3(2x+1) when x = 4.

Solution

Plug x = 4 in the function f(x).

f (4) = 3[2(4) + 1]

f (4) = 3[8 + 1]

f (4) = 3 x 9

f (4) = 27

Example 3

Write the function y = 2x2 + 4x – 3 in function notation and find f (2a + 3).

Solution

y = 2x2 + 4x – 3 ⟹ f (x) = 2x2 + 4x – 3

Substitute x with (2a + 3).

f (2a + 3) = 2(2a + 3)2 + 4(2a + 3) – 3

= 2(4a2 + 12a + 9) + 8a + 12 – 3
= 8a2 + 24a + 18 + 8a + 12 – 3
= 8a2 + 32a + 27

3 0

2 years ago

Solve for x: x^2-2x=15

Lyrx [107]

Answer:

Step-by-step explanation:

x^2 - 2x = 15

x^2 - 2x - 15 = 15 - 15 (Subtract 15 from both sides.)

x^2 - 2x -15 = 0 (Simplify.)

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<em>Set up & solve for a quadratic formula</em>

<h3><em>* I will solve for your first x</em></h3>

2 + root 64 / 2x 1 = 2 + root 64 / 2

root 64 = 8

= 2 + 8 / 2

= 10 / 2

= 5

<em>To solve for the second, follow the same pattern/formula for the first that I used to get 5.</em>

5 0

2 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><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)

3 0

8 months ago