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3 years ago

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What is Jaime’s speed if he ran 12 miles in h hours? (He was running with the same speed all that time, believe it or not.) PLS

HELP ASAP

2 answers:

Alexandra [31]3 years ago

7 0

Answer:

speed= time/distance

speed=h/12

step-by-step explanation:

likoan [24]3 years ago

4 0

Answer: The answer is 12/h

Step-by-step explanation:

distance/ speed

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A woman a walked 12km at 3km/hr and cycled 18km at 9km/hr. what is her average speed for the whole journey.

Deffense [45]

Answer:

5 km/h

Step-by-step explanation:

7 0

2 years ago

On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

Answer:

a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

7 0

2 years ago

What and which at time are the maximum and minimum speeds in this function?

Svetach [21]

Answer:

the maximum is 12000 and the minimum is 0

Step-by-step explanation:

sorry if this doesn't answer your question

4 0

2 years ago

Read 2 more answers

It is important to know the appropriate dose of ibuprofen for a child so that you are ready in the case of high fever. The dosin

Alexus [3.1K]

Answer:

102 mg

Step-by-step explanation:

Let's convert 28 pounds to kgs.

Since 1 kg = 2.2 pounds, we can find:

$\frac{1Kg}{2.2Pounds}=\frac{xKg}{28Pounds}\\1*28=2.2x\\28=2.2x\\x=\frac{28}{2.2}\\x=12.73$

So 28 pounds is 12.73 Kg

Since 8mg is given per kg, for 12.73 kg, the amount given should be:

12.73 * 8 = 101.84

Rounded to nearest whole number, that is, 102 mg

8 0

3 years ago

Read 2 more answers

Help me please i don't know what to do

astra-53 [7]

Answer:

a. a = 1, b = -5, c = -14

b. a = 1, b = -6, c = 9

c. a = -1, b = -1, c = -3

d. a = 1, b = 0, c = -1

e. a = 1, b = 0, c = -3

Step-by-step explanation:

a. x-ints at 7 and -2

this means that our quadratic equation must factor to:

$(x-7)(x+2) = 0$

FOIL:

$x^2 + 2x - 7x - 14 = 0$

Simplify:

$x^2 - 5x - 14 = 0$

a = 1, b = -5, c = -14

b. one x-int at 3

this means that the equation will factor to:

$(x-3)^2=0$

FOIL:

$x^2 - 3x - 3x +9 = 0$

Simplify:

$x^2 - 6x + 9 = 0$

a = 1, b = -6, c = 9

c. no x-int and negative y must be less than 0

This means that our vertex must be below the x-axis and our parabola must point down

There are many equations for this, but one could be:

$-x^2-x-3=0$

a = -1, b = -1, c = -3

d. one positive x-int, one negative x-int

We can use any x-intercepts, so let's just use -1 and 1

The equation will factor to:

$(x+1)(x-1)=0$

This is a perfect square

FOIL:

$x^2-1=0$

a = 1, b = 0, c = -1

e. x-int at $\sqrt{3} , -\sqrt{3}$

our equation will factor to:

$(x+\sqrt{3} )(x-\sqrt{3)} =0$

This is also a perfect square

FOIL and you will get:

$x^2 - 3 = 0$

a = 1, b = 0, c = -3

4 0

1 year ago