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2 years ago

6

A rhombus, which is a quadrilateral with 44 equal sides, is plotted on the coordinate plane. The coordinates of the vertices are

: (5,3)5,3, (2,-1)2,-1, (-3,-1)-3,-1, (0,3)0,3. How long is each side of the rhombus?

1 answer:

Arturiano [62]2 years ago

8 0

The rhombus is shown in the diagram below

The length of one side of the rhombus is 5 units, hence all the other sides are also 5 units each.

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3 years ago

Need help asap plz help dont have many points but ill give yu all i have if u help

Anna71 [15]

Hello!

We can go through the choices and see if they are true

Choice 1
A function has one output for every input

This is not a function because it has -1 as an output for both 4 and -1

This is the answer
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Choice 2
The inputs are the x's

We see that the inputs are -1, 0, 3, and 4

This is true
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Choice 3
The outputs are the y's

We see the outputs are -2, -1, and 1

This is true
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Choice 4
We can follow the lines and see the relations are

(-1, 1), (0, -2), (3, 1), (4, -1)

This is also true
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The only one that was false was the first choice so that is the answer

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4 0

2 years ago

the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i

nekit [7.7K]

Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

$A = \frac{1}{2}\cdot b\cdot h$ (Eq. 1)

Where:

$A$ - Area of the triangle, measured in square centimeters.

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

$\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt}$ (Eq. 2)

Where:

$\frac{dA}{dt}$ - Rate of change of area in time, measured in square centimeters per minute.

$\frac{db}{dt}$ - Rate of change of base in time, measured in centimeters per minute.

$\frac{dh}{dt}$ - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

$\frac{1}{2}\cdot\frac{db}{dt}\cdot h = \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}$

$\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt}$ (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

$b = \frac{2\cdot A}{h}$

If we know that $A = 130\,cm^{2}$ , $h = 15\,cm$ , $\frac{dh}{dt} = 2.5\,\frac{cm}{min}$ and $\frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}$ , the rate of change of the base of the triangle in time is:

$b = \frac{2\cdot (130\,cm^{2})}{15\,cm}$

$b = 17.333\,cm$

$\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)$

$\frac{db}{dt} = -2.262\,\frac{cm}{min}$

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

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2 years ago