Nal or irrational Classify each number below as a rational number or an irrational number.

At 3:30 p.m., Berto’s train was 34 miles past the egg farm, traveling at an average speed of 85 miles per hour. At the same time

timurjin [86]

E=egg farm

x=distantce between the Berto´s trains and the meeting point.

t=time where the trains meet up

⇒110 miles/h ⇒85miles/h

[---------------------------E----------------------]-------------------------------------X

[............12 miles.......][........34 miles.....][------------ x ----------------------]

distance between Berto´s train and Eduardo´s train=

=12 miles+34 miles=46 miles.

S=d/t ⇒d=S*t

S=seed

d=distance

T=time

Eduardo´s train;

46 miles+x=t*110 miles/h ⇒x=t*110 miles/h-46 miles (1)

Berto´s train;

x=t*85 miles/h (2)

With the equations (1) and (2) we suggest this system equations:

x=t* 110 miles/h-46 miles

x=t*85 miles/h

we solve this system equiations :

t*110 miles/h-46 miles=t*85 miles/h

110t miles/h-85t miles/h=46 miles

25 t miles/h=46 miles

t=46 miles/25 miles/h=1,84 hour (≈1 hour, 50 minutes, 24 seconds)

x=t*85 miles/h=156,4 miles

Solution: 1,84 hour

3 0

3 years ago

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A rocket is fired with an initial vertical velocity of 40 m/s from a launch pad 45 m high, and its height is given by the formul

N76 [4]

Answer:

1) 125 meters

2) For 9 seconds.

Step-by-step explanation:

The rocket's height is given by the formula $-5t^2+40t+45$ .

Notice that this is a quadratic.

Part 1)

Since this is a quadratic, the highest height the rocket goes will be the vertex of our quadratic. Remember that the vertex of a quadratic in standard form is:

$(-\frac{b}{2a}, f(-\frac{b}{2a}))$

So, let's identify our coefficients. The standard quadratic form is:

$ax^2+bx+c$

Therefore, in this case, our a is -5, b is 40, and c is 45.

So, let's find the x-coordinate of our vertex. Substitute 40 for b and -5 for a. This yields:

$x=\frac{-(40)}{2(-5)}$

Multiply and reduce. So:

$x=-40/-10=4$

Now, substitute 4 for t to find the height. So:

$-5(4)^2+40(4)+45$

Evaluate:

$=-5(16)+40(4)+45$

Multiply and add:

$=-80+160+45=125\text{ meters}$

Therefore, the highest the rocket goes up is 125 meters.

Part 2)

To find out for how much time the rocket is in the air, we can think about after how many seconds after launch will the rocket land.

If the rocket lands, the height h will be 0. So, we can set our expression equal to 0 and solve for t:

$-5t^2+40t+45=0$

First, let's divide everything by -5. This yields:

$t^2-8t-9=0$

We can factor:

$(t+1)(t-9)=0$

Zero Product Property:

$t+1=0\text{ or } t-9=0$

Solve for t:

$t=-1\text{ or } t=9$

In this case, since t is our time in seconds, -1 seconds does not make sense. So, we can remove -1 from our solution set.

Therefore, 9 seconds after launch, the rocket will touch the ground.

Therefore, the rocket was in the air for 9 seconds.

And we're done!

6 0

3 years ago

Acellus

steposvetlana [31]

Format: y = mx + b
m = slope, b = y intercept
The answer is y = x + 2

4 0

3 years ago

The perimeter of a playing field for a certain sport is 334ft. The field is a rectangle, and the length is 47 ft longer than the

babunello [35]

Answer:

The dimensions of the rectangle = 60ft by 107ft

Where 60 ft = Width of the playing field

107ft = Length of the playing field

Step-by-step explanation:

A playing field is Rectangular is shape, hence,

The formula for Perimeter of a rectangle = 2(L + W)

P = 334 ft

L = 47 + W

W = W

Hence we input these values in the formula and we have:

334 = 2(47 + W + W)

334 = 2(47 + 2W)

334 = 94 + 4W

334 - 94 = 4W

240 = 4W

W = 240/4

W = 60

There fore, the width of this playing field = 60 ft

The length of this rectangle is calculated as:

47 + W

47 + 60

= 107 ft

The length of this playing field = 107ft

Therefore the dimensions of the rectangle = 60ft by 107ft

6 0

2 years ago

Solve with solution...no spam answers pls

Agata [3.3K]

Let's analyze Hannah's work, step-by-step, to see if she made any mistakes.

In Step 1, Hannah wrote $\dfrac{d}{dx} (-3+8x)$ as the sum of two separate derivatives $\dfrac{d}{dx}(-3)+ \dfrac{d}{dx} (8x)$ using the sum rule.

This step is perfectly fine.

In Step 2, $\dfrac{d}{dx}(8x)$ was kept as it is, and $\dfrac{d}{dx}(-3)$ was rewritten as $0$ using the constant rule.Indeed, according to the constant rule, the derivative of a constant number is equal to zero.

This step is perfectly fine.

In Step 3, $\dfrac{d}{dx} (8x)$ was rewritten as $\dfrac{d}{dx}(8) \dfrac{d}{dx}(x)$ supposedly using the constant multiple rule.

The problem is that according to the constant multiple rule, $\dfrac{d}{dx}(8x)
$ should be rewritten as $8 \dfrac{d}{dx}(x)$ and not as $\dfrac{d}{dx}(8)\dfrac{d}{dx}(x)$ .

Therefore, Hannah made a mistake in this step.

6 0

3 years ago

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