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3 years ago

Pls fast,i need it ..............................

Mathematics

1 answer:

Ghella [55]3 years ago

6 0

Answer:

hints to solve each. partial answer due to lack of time, hope it helps!

Step-by-step explanation:

a) solve ():

5 $\sqrt{3}$ + 7 $\sqrt{3}$ - 9 $\sqrt{3}$ - 6 $\sqrt{3}$ + 12 $\sqrt{3}$ - 10 $\sqrt{3}$ =

$\sqrt{3}$ . (5 +7-9-6+12-10)

b) use $\sqrt{6\\}$ = $\sqrt{3\\}\\\\$ $\sqrt{2\\}\\\\$

c) use $\sqrt{20\\}$ = $\sqrt{4\\}\\\\$ $\sqrt{5\\}\\\\$ = 2 $\sqrt{5\\}\\\\$

use $\sqrt{8\\}$ = $\sqrt{4\\}\\\\$ $\sqrt{2\\}\\\\$ = 2 $\sqrt{2\\}\\\\$

a) use $\sqrt{30\\}$ = $\sqrt{6}$ $\sqrt{5}$

use $\sqrt{30\\}$ = $\sqrt{6}$ $\sqrt{4}$ = 2 $\sqrt{6}$

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Zepler [3.9K]

Answer:

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Step-by-step explanation:

7 0

3 years ago

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

What is the perimeter of a squre window with sides 36 inches long ?

Solnce55 [7]

144 inches .....
hope this helps

4 0

3 years ago

Read 2 more answers

Why do the factor not have to be prime?

dangina [55]

By definition of prime! An integer greater than one is called a prime number if its only positive divisors (factors) are one and itself.

5 0

4 years ago

Read 2 more answers

Health insurance benefits vary by the size of the company. The sample data below show the number of companies providing health i

telo118 [61]

Answer:

Step-by-step explanation:

a. The hypothesis of testing are

Null Hypothesis, H0: The health insurance coverage is independent of the size of the company.

Alternative hypothesis, H1: The health insurance coverage is not independent of the size of the company

The test statistics is calculated as

X² = ∑(Ei - Oi)² ≅X²(r-1)(c-1)

i Ei

where Oi's are the observed values and Ei's are the expected values,

E= row total × column total

total

The expected values are

Sizeof the Company

Health Insurance Small Medium Large Total

Yes 42 63 84 189

No 8 12 16 36

Total 50 75 100 225

Therefore, the test statistics is

X² =∑ (Ei -Oi)² =6.94 ≅ x²₂

i Ei

The critical value is found to be 5.99. Since test statistics > critical value, we reject H0 and conclude that the health insurance coverage is not independent of the size of the company.

The associated p-value is 0.031 (<0.05).

b. Percentage of employees lacking health insurance coverage is

Small 14/50 x 100 = 28%

Medium 10/75 x 100 = 13%

Large 12/100 x 100 = 12%

Hence depending on the above percentage, we can support the claim that the employees of small companies are more likely to lack health insurance coverage.

8 0

4 years ago