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3 years ago

5

A rational number subtracted by a rational number equals?

1 answer:

zimovet [89]3 years ago

4 0

You still get a rational number.

The sum/subtraction between rational numbers is defined as follows:

$\dfrac{a}{b}\pm\dfrac{c}{d} = \dfrac{ad\pm bc}{bd}$

The expression $ad\pm bc$ is the sum/difference between the products of integers, and so is still an integer. This means that

$\dfrac{ad\pm bc}{bd}$

is still a ratio between integers, and thus a rational.

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PLEASE SOMEONE HELP! Thanksss

marusya05 [52]

Answer:

Ф=144

Ф =s/r ( Ф is the central angle,s is the arc length, r is the radius)

find r:

radius =circumference/2π=1/2π=0.16

Ф=(2/5) /(1/2π)=4π/5=144 degree (π=180)

Ф in degrees: 144 degrees

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3 years ago

Find the measure of an exterior angle of a regular polygon with 9 sides

Contact [7]

Answer: 40°

Step-by-step explanation: The formula for calculating the size of an exterior angle is: exterior angle of a polygon = 360 ÷ number of sides. So it's 360°÷9=40°

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3 years ago

Its number 1. 10 Points to answer

Pani-rosa [81]

Answer : 21 =y3
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3 years ago

Read 2 more answers

22.5(10-y)+27.5y=250

Nataliya [291]

Answer:

y = 5

Step-by-step explanation:

22.5(10-y)+27.5y=250

$225 - 22.5y + 27.5y = 250\\\\$

Combind like terms,

$225 + 5y = 250$

Subtract 225 from both equations to isolate the variable.

$5y = 25$

Divide both sides by 5 to isolate the variable and solve it.

$y = 5$

4 0

3 years ago

PLEASE HELP!!! WILL MARK BRAINLIEST!! THX! A golfer hits a ball with an initial velocity of 32.7 m/s from the ground. Find the f

OLEGan [10]

Answer:

See below

Step-by-step explanation:

<u>First Problem</u>

The ball hits the ground when $h(t)=0$ , therefore:

$h(t)=-4.9t^2+v_0t+h_0$

$0=-4.9t^2+32.7t$

$0=t(-4.9t+32.7)$

$t=0$ and $t=\frac{32.7}{4.9}\approx6.67$

Since the ball is in the air before it hits the ground, $t=6.67$ (seconds) is the more appropriate choice.

<u>Second Problem</u>

The maximum height of the ball is determined when $t=-\frac{b}{2a}$ , therefore:

$t=-\frac{b}{2a}$

$t=-\frac{32.7}{2(-4.9)}$

$t=-\frac{32.7}{-9.8}$

$t\approx3.34$

This means that the height of the ball is at its maximum after 3.34 seconds:

$h(t)=-4.9t^2+32.7t$

$h(3.34)=-4.9(3.34)^2+32.7(3.34)$

$h(3.34)\approx54.55$

Thus, the answer is 54.55 (meters).

<u>Third Problem</u>

Refer to the second problem

<u>Fourth Problem</u>

<u /> $h(t)=-4.9t^2+32.7t$ <u />

<u /> $h(4.3)=-4.9(4.3)^2+32.7(4.3)$ <u />

<u /> $h(4.3)\approx50.01$ <u />

<u />

Therefore, the height of the ball after 4.3 seconds is 50.01 (meters).

<u>Fifth Problem</u>

The ball will be 24 meters off the ground when $h(t)=24$ , therefore:

$h(t)=-4.9t^2+32.7t$

$24=-4.9t^2+32.7t$

$0=-4.9t^2+32.7t-24$

$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$t=\frac{-32.7\pm\sqrt{(32.7)^2-4(-4.9)(-24)}}{2(-4.9)}$

$t_1\approx0.84$

$t_2\approx5.83$

Therefore, the ball will be 24 meters off the ground after 0.84 (seconds) and 5.83 (seconds)

5 0

2 years ago