Two fifths divided by one half times ( negative one minus 2)

A number c minus 3.4 is less than 14. Write this word sentence as an inequality.

Rufina [12.5K]

C - 3.4 < 14 <=== ur inequality

3 0

3 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

3 years ago

What is the median of the data set?

olga2289 [7]

<h3>Answer: 8.5 (choice B)</h3>

=========================================================

Explanation:

The diagram shows us we have this data set

{5, 8, 8, 8, 8, 9, 9, 9, 10, 10}

Cross off the outer most items (the 5 and the second 10) to get this smaller set

{8, 8, 8, 8, 9, 9, 9, 10}

Repeat the last step of crossing off the outer most items to get

{8, 8, 8, 9, 9, 9}

Keep this process going until we have this set of steps

{8, 8, 9, 9}

{8, 9}

We have two items left, which are tied for the middle, so we compute the midpoint

(8+9)/2 = 17/2 = 8.5 is the median

This is the middle most value

7 0

3 years ago

Explain how an equation is like a balanced hanger plzzz answer due today

NikAS [45]

A hanger stays balanced when the weights on both sides are equal.

An equation can be compared to a balanced hanger. We can change the equation, but for a true equation to remain true, the same thing must be done to both sides of the equal sign.

8 0

3 years ago

A store that buys and sells used video games buys a game at a cost of $9 and sells it at a selling price of $12. Find the ma

d1i1m1o1n [39]

Answer:

a) markup = $3

b) markup as a percentage of cost is 33.3%

Step-by-step explanation:

<em>Markup</em>

... markup = selling price - cost

... = $12 - 9

... markup = $3

<em>Markup as a Percentage of Cost</em>

To find the percent markup, divide the markup by the reference value and multiply the ratio by 100%. The reference value for markup is usually cost price, but sometimes may be selling price.

... markup / cost × 100% = 3/9×100% = 33 1/3% ≈ 33.3%

_____

When the markup is figured as a percentage of selling price, the result is usually called <em>margin</em>. Some folks are not very careful in distinguishing markup from margin.

3 0

3 years ago