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3 years ago

Explain in your own words, what are reciprocals and when are they used?

2 answers:

7 0

Recipirocals is 1 divided by the number. The number times the reciprocal equals 1. Reciprocals are used when trying to get something to equal one, like when clearing out a fraction.

photoshop1234 [79]3 years ago

4 0

They are multiplicative inverses. For example, the reciprocal of 2/3 is 3/2. They are used when dividing fractions

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3. A rocket is launched from a height of 3 meters with an initial velocity of 15 meters per second.

Vikki [24]

Let the rocket is launched at an angle of \theta with respect to the positive direction of the x-axis with an initial velocity u=15m/s.

Let the initial position of the rocked is at the origin of the cartesian coordinate system where the illustrative path of the rochet has been shown in the figure.

As per assumed sine convention, the physical quantities like displacement, velocity, acceleration, have been taken positively in the positive direction of the x and y-axis.

Let the point P(x,y) be the position of the rocket at any time instant t as shown.

Gravitational force is acting downward, so it will not change the horizontal component of the initial velocity, i.e. $U_x=U cos\theta$ is constant.

So, after time, t, the horizontal component of the position of the rocket is

$x= U \cos\theta t \;\cdots(i)$

The vertical component of the velocity will vary as per the equation of laws of motion,

$s=ut+\frac12at^2\;\cdots(ii)$ , where,s, u and a are the displacement, initial velocity, and acceleration of the object in the same direction.

(a) At instant position P(x,y):

The vertical component of the initial velocity is, $U_y=U sin\theta$ .

Vertical displacement =y

So, s=y

Acceleration due to gravity is $g=9.81 m/s^2$ in the downward direction.

So, $a=-g=-9.81 m/s^2$ (as per sigh convention)

Now, from equation (ii),

$y=U sin\theta t +\frac 12 (-9.81)t^2$

$\Rightarrow y=U \sin\theta \times \frac {x}{U \cos\theta} +\frac 12 (-g)\times \left(\frac {x}{U \cos\theta} \right)^2$

$\Rightarrow y=U^2 \tan\theta-\frac 1 2g U^2 \sec^2 \theta\;\cdots(iii)$

This is the required, quadratic equation, where $U=15 m/s$ and $g=9.81 m/s^2$ .

(b) At the highest point the vertical velocity,v, of the rocket becomes zero.

From the law of motion, $v=u+at$

$\Rightarroe 0=U\sin\theta-gt$

$\Rightarroe t=\frac{U\sin\theta}{g}\cdots(iv)$

The rocket will reach the maximum height at $t= 1.53 \sin\theta s$

So, from equations (ii) and (iv), the maximum height, $y_m$ is

$y_m=U\sin\theta\times \frac{U\sin\theta}{g}-\frac 12 g \left(\frac{U\sin\theta}{g}\right)^2$

$\Rightarrow y_m=23 \sin\theta -11.5 \sin^2\theta$

In the case of vertical launch, $\theta=90^{\circ}$ , and

$\Rightarrow y_m=11.5 m$ and $t=1.53 s$ .

Height from the ground= 11.5+3=14.5 m.

(c) Height of rocket after t=4 s:

$y=15 \sin\theta \times 4- \frac 12 (9.81)\times 4^2$

$\Rightarrow y=15 \sin\theta-78.48$

$\Rightarrow -63.48 m >y> 78.48$

This is the mathematical position of the graph shown which is below ground but there is the ground at $y=-3m$ , so the rocket will be at the ground at t=4 s.

(d) The position of the ground is, y=-3m.

$-3=U\sin\theta t-\frac 1 2 g t^2$

$\Rightarrow 4.9 t^2-15 \sin\theta t-3=0$

Solving this for a vertical launch.

$t=3.25 s$ and $t=-0.19 s$ (neglecting the negative time)

So, the time to reach the ground is 3.25 s.

(e) Height from the ground is 13m, so, y=13-3=10 m

$10=U\sin\theta t-\frac 1 2 g t^2$

Assume vertical launch,

$4.9 t^2-15 \sin\theta t+10=0$ [using equation (ii)]

$\Rightarrow t=2.08 s$ and $t=0.98 s$

There are two times, one is when the rocket going upward and the other is when coming downward.

4 0

3 years ago

Decribe the counter clockwise rotation that maps angles

Llana [10]

I don’t know if this is the answer but.. An angle generated by one complete counterclockwise rotation measures 360° or 21 radians. 360° An angle generated by one complete clockwise rotation measures -360° or -21 radians.

5 0

3 years ago

I am confused on numbers 25 and 29, the instructions are at the top.

kvasek [131]

#25 is fairly simple. Plug in -4 and 3 into the equation, and the extraneous root will be the one that does not work.
$\sqrt{12-(-4)} = \sqrt{16} = \frac{+}{}4$
Extraneous root in this case is positive four since +4≠-4
 $\sqrt{12-3} = \sqrt{9} = \frac{+}{}3$
In this case it's negative 3, since -3≠3

#29 can be turned into a quadratic equation.
$x= \sqrt{2x+3}$
Square both sides to get
$x^{2}=2x+3$
Then bring the 2x+3 to the other side, setting the quadratic equal to zero.
$x^{2}-2x-3=0$
Factor to find that it's equivalent to
(x-3)(x+1)=0
Therefore x is equal to positive 3 and negative 1. Plug both back into the original equation. Whichever does not work is the extraneous root, and the answer is the one that does.
 $x= \sqrt{2x+3}$
 $3= \sqrt{2(3)+3}$
 $3= \sqrt{9}$
Extraneous root would be negative 3.

 $-1= \sqrt{2(-1)+3}$
 $-1= \sqrt{1}$
Extraneous root would be positive 1.

Your answers are positive 3 and negative 1.
Extraneous roots are negative 3 and positive 1.

3 0

3 years ago

While shopping for a speaker, you see a discount.

harina [27]

Answer:

3*4. and 0.3*42.00 and 0.03*42

Step-by-step explanation:

4 0

3 years ago

Write the equation of the graph in slope-intercept form y=mx+b

s344n2d4d5 [400]

Answer:

y=(2/5)x-1

Step-by-step explanation:

okay so basically just take the Y intercept, so (-1,0) and put it in for b, then find the slope, which is 2/5, and plug it in for m.

4 0

3 years ago