What is 1600+3200 pls

12 times 1/8 how do you cross multiply this

ra1l [238]

Answer:

12/1 times 1/8 = 1.5

Step-by-step explanation:

You multiply 12 by 1 to get 12, then divide 12 by 8 to get 1.5

7 0

3 years ago

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Under her cell phone plan, Autumn pays a flat cost of $48 per month and $5 per gigabyte. She wants to keep her bill at $68.50 pe

VikaD [51]

Answer:

4 gigabytes

Step-by-step explanation:

Equation: 68.50=48+5x

8 0

2 years ago

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a pair of sunglasses is on sale for $26. The original price of the sunglasses was $40. What percent of the original price is the

Xelga [282]

65%...
26 divided by 40 = 0.65 x 100 = 65%
hope this helps!

5 0

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>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:

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