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3 years ago

Simplify x^4y^3 over x^2y^5

Mathematics

2 answers:

jek_recluse [69]3 years ago

4 0

$\dfrac{x^4y^3}{x^2y^5} =$

$= \dfrac{x^2}{y^2}$

kondaur [170]3 years ago

4 0

If you mean (x^4 * y^3) over (x^2 * y^5) it simplifies to

x^2/y^2

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Which expression is equivalent to 2x2 – 2x + 7?

diamong [38]

11-2x would be the answer u are looking for. Hope this helps!

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4 years ago

Read 2 more answers

The students in Mr. Wilson's Physics class are making golf ball catapults. The

Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is $0 \leq x \leq 48.6\ ft$ and the range is $0 \leq y \leq 8.3\ ft$

Step-by-step explanation:

we have

$y=-0.014x^{2} +0.68x$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's distance from the catapult in feet

y is the flight of the balls in feet

Part a) How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-0.014x^{2} +0.68x=0$

$a=-0.014\\b=0.68\\c=0$

substitute in the formula

$x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}$

$x=\frac{-0.68(+/-)0.68} {(-0.028)}$

$x=\frac{-0.68(+)0.68} {(-0.028)}=0$

$x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft$

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

$y=-0.014x^{2} +0.68x$

Find out the derivative and equate to zero

$0=-0.028x +0.68$

Solve for x

$0.028x=0.68$

$x=24.3$

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint between the x-intercepts

$x=(0+48.6)/2=24.3\ ft$

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

$y=-0.014(24.3)^{2} +0.68(24.3)$

$y=8.3\ ft$

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A reasonable domain is the distance between the two x-intercepts

$0 \leq x \leq 48.6\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

we have the interval -----> [0,8.3]

$0 \leq y \leq 8.3\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

8 0

3 years ago

The diagonal shortcut across a rectangular parking lot is 169 feet long. The lot is 144 feet wide. What is the length of the lot

Alexeev081 [22]

Answer: I think it is 25.5ft

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3 years ago

Which conversion factors can be used to find the number of hours in a week? Select all that apply.

Archy [21]

Answer:

3rd and 4th options are correct that is 24 hours per day and 7 days per week.

Step-by-step explanation:

We need to find the conversion factors that can be used to find number of hours in a week.

We know that,

Number of hours in a day = 24

and

Number of days in a week = 7

So, Number of hours in 7 days = 24 × 7 = 168.

Therefore, 3rd and 4th options are correct that is 24 hours per day and 7 days per week.

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3 years ago

Solve the equation on the interval [0, 2π). <br> 3(sec x)^2 -4=0

Flauer [41]

$\: \: \: \: \:$

$x = \frac{\pi}{6} , \frac{(k)\pi}{6} , \frac{(5)\pi}{6} , \frac{(k)\pi}{6}$

<h3><em>refer</em><em> to</em><em> the</em><em> attachment</em></h3><h2><u>hope</u><u> it</u><u> helps</u></h2><h2 />

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3 years ago