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3 years ago

Answer ASAP, will give brainliest

Mathematics

1 answer:

Bond [772]3 years ago

8 0

Answer:

Step-by-step explanation:

1) Diagonal bisect the angles of Rhombus

∠CAB = ∠CAD

∠CAB = 71

2) ∠DAB = ∠CAB + ∠CAD

∠DAB = 71 + 71 = 142

In Rhombus, adjacent angles are supplementary

∠DAB + ∠ABC = 180

142 + ∠ABC = 180

∠ABC = 180 - 142

∠ABC = 38

3) In rhombus, opposite angles are congruent

∠ADC = ∠ABC

∠ABC = 38

In rhombus, diagonal bisect angles

∠BDC = (1/2)*∠ADC

∠BDC= 38/2

∠BDC = 19

4) Diagonals bisect each other at 90

∠DEC = 90

5) Diagonals bisect each other

BE = DE

BE + DE = DB

7x -2 +7x -2 =24 {add like terms}

14x - 4 =24

14x = 24+4

14x = 28

x = 28/14

x = 2 m

6) AB = 13m

BE = 7x - 2 = 7*2 -2 = 14 -2 = 12 m

In right angle ΔAEB, {use Pythagorean theorem}

AE² + BE² = AB²

AE² + 12² = 13²

AE² + 144 = 169

AE² = 169 - 144

AE² = 25

AE = √25

AE = 5 m

Diagonals bisect each other

AE = EC

AC = 2*5

AC = 10 m

7)Side = 13 m

Perimeter = 4*side

= 4*13

Perimeter = 52 m

8) d1 = AC = 10 m

d2 = DB = 24 m

Area = $\frac{d_{1}*d_{2}}{2}$

$=\frac{10*24}{2}\\$

= 10 *12

= 120 m²

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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.