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3 years ago

What is the perimeter of rhombus wxyz?

Mathematics

2 answers:

Lunna [17]3 years ago

5 0

For this case, the first thing we must do is define the formula of distance between points.

We have then:

$d = \sqrt{(x2-x1)^2 + (y2-y1)^2}$

From here, we look for the distance between two points of rhombus.

$YZ = \sqrt{(5-3)^2 + (5-2)^2} \\YZ = \sqrt{2^2 + 3^2} \\YZ = \sqrt{4 + 9} \\YZ = \sqrt{13}$

Then, since all sides have the same length, then the perimeter is given by:

$P = 4YZ$

Substituting we have:

$P = 4\sqrt{13}$

Answer:

The perimeter of the rhombus is:

$P = 4\sqrt{13}$

option 3

VikaD [51]3 years ago

4 0

One of the properties of rhombus is its sides are congruent. Therefore, we will only need to find the length of one side using the distance formula. At line YZ, the distance is $\sqrt{ (x_{2} -x_{1} )^{2} +(y_{2} -y_{1} )^{2}}$

D = $\sqrt{ (5 -3 )^{2} +(5 -2) ^{2} }$ = √13

To get the perimeter, multiply one side to 4. Hence, the perimeter is 4√13 (C)

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2 years ago

A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.

sattari [20]

Answer:

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

$y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$y_{o}$ - Initial height of the ball, measured in feet.

$v_{o}$ - Initial speed of the ball, measured in feet per second.

$g$ - Gravitational constant, equal to $-32.174\,\frac{ft}{s^{2}}$ .

$t$ - Time, measured in seconds.

Given that $y_{o} = 0\,ft$ , $v_{o} = 95\,\frac{ft}{s}$ , $g = -32.174\,\frac{ft}{s^{2}}$ and $y = 120\,ft$ , the following second-order polynomial is found:

$120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}$

$-16.087\cdot t^{2} + 95\cdot t -120 =0$

The roots of this polynomial are, respectively:

$t_{1} \approx 4.075\,s$ and $t_{2} \approx 1.831\,s$ .

Both roots solutions are physically reasonable, since $t_{1}$ represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas $t_{2}$ represents the instant when the ball the same height after reaching maximum height.