What the answer for x and y

(4k – 2) (k + 4)<br> O 44² + 14k - 8<br> O 4K² - 8<br> O 5k + 2<br> 0 42² + 16k - 8

andreyandreev [35.5K]

Answer:

4k^2 + 14k - 8

Step-by-step explanation:

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below ÷)

4 0

3 years ago

Find the value of x.<br><br> (x + 4)<br> 147°

larisa86 [58]

Answer:

Step-by-step explanation:

From the question the angles lie on a straight line which means that the sum of their angles add up to 180°

Since angles on a straight line add up to 180°

To find the value of x add up all the angles and equate it to 180 to find x

That's

x + 4 + 147 = 180

x + 151 = 180

x = 180 - 151

We have the final answer as

Hope this helps you

8 0

4 years ago

What would be the volume of the pyramid? Little help please

murzikaleks [220]

Answer:

36

Step-by-step explanation:

The answer is 36, I did the work

7 0

3 years ago

A rectangle is fenced off for a robot competition. The dimensions of the competition area are 35 feet by 25 feet. To help track

xz_007 [3.2K]

Answer:

The approximate speed of the robot during the 5 seconds is:

<u>6 ft/s</u>

Step-by-step explanation:

To calculate the speed of the robot, you must begin with the positions, the first position (3, 18) and the second position (31, 6), you can see, in the height it moved from 3 to 31, it means 28 feet, in the width it moves from 18 to 6, it means 12 feet, with these data you can construct a triangle where you have the opposite leg and adjacent leg, now you must calculate the hypotenuse, because it is the linear path from the first position to the second position that the robot took, for this, you can use the Pythagoras theorem:

$hypotenuse^{2}$ = $opposite leg^{2}$ + $adjacent leg^{2}$
$hypotenuse^{2}$ = $12^{2} +28^{2}$
$hypotenuse=\sqrt{ 12^{2}+28^{2} }$
$hypotenuse=30.46 ft$

With the value of the distance traveled, and the time used (5 seconds), we can calculate the speed with the next formula:

$Speed =\frac{distance}{time}$
$Speed = \frac{30.46 ft}{5 s}$
Speed = 6.09 ft/s

As you need the speed in the nearest whole number, then:

<u>Speed = 6 ft/s</u>

5 0

3 years ago

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .

7 0

3 years ago