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3 years ago

A line intersects the y-axis of the point (a, b) Is a = 0? Is b = 0? Explain.

Mathematics

1 answer:

Ber [7]3 years ago

7 0

9514 1404 393

Answer:

a=0

b may be 0 but is not required to be

Step-by-step explanation:

The y-axis corresponds to the line x=0. Any point on the y-axis will have an x-coordinate of 0.

If (a, b) is on the y-axis, then a = 0.

If the intersection point is (0, 0), then b=0. Otherwise, b is not 0.

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Saj has made 7 litres of curry sauce.

Makovka662 [10]

Answer:

18 jars can be filled completely.

Step-by-step explanation:

7L=7000ml

1 jar contain 375 ml

375ml can be filled in 1 jar

375ml=1 jar

1 ml = 1/375 jar

7000 ml = 7000× 1/375 jar

7000ml= 18.67 jar

hence 18 jars can be filled completely.

3 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

Would you please help me solve for x it should be a degrees but I’m not sure what it is

Leno4ka [110]

Answer:

x=15º

Step-by-step explanation:

A triangle's angles add up to 180º.

∠A=30º

∠C=180º-45º

∠B=xº

To find ∠B we have to first find ∠C.

Since ∠C does not have an angle, knowing that a straight line is 180º we can subtract it from 45º to find ∠C.

180-45=135

∠C=135º

Now add ∠A and ∠C.

30+135=165

Subtract the sum from 180º

180-165=x

x=15º

Hope this helps :)

6 0

3 years ago

How do you write 36 as a product of two factors in two different ways?

weeeeeb [17]

<span>6 x 6 = 36 it is the square root of 36
2 x 18 = 36 it is a multiple of 2
4 x 9 = 36 it is a multiple of 4
3 x 12 = 36 it is a multiple of 3
Hope this helps a bit
happy to help you
</span>

7 0

3 years ago

Using the digits 3, 4, 7, and 8, write the largest possible number in word form, if 3 is in the hundreds place.

blsea [12.9K]

Biggest number would be 8743, but 3 must be hundreds so the biggest will be 8374. Which is eight thousand three hundred seventy four

5 0

3 years ago

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