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3 years ago

13

I am a quadrilateral with 2 pairs of opposite sides that are parallel, 2 pairs of sides that are of equal length,and 4 right ang

les. I am not a square. What shape am I?

1 answer:

Ludmilka [50]3 years ago

5 0

Answer:

Rectangle

Step-by-step explanation:

The shape is of a rectangle who has got two pairs of opposite sides that are parallel and two pairs of sides that are of equal length of each other with all the four right angles that makes up total angle of 360 degrees in total. It can be acquired that the area which is equal to length multiply by breadth.

Calculation,

A length of the rectangle is equal to each other,

Assuming it to be 3cm,

Breadth of it is equal to each other,

Assuming that to be 2cm,

Length x Breadth = Area

3 x 2 = 6cm to the power 2.

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Read 2 more answers

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stealth61 [152]

Answer:

11

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3 years ago

Is b = 6 a solution to the inequality b – 3 > 5?

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Step-by-step explanation:

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2 years ago

Write the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10).

Serjik [45]

Answer:

$\displaystyle f(x)=x^2+2x+2$

Step-by-step explanation:

<u>System Of Linear Equations </u>

In this problem, we'll need to solve a 3x3 system of linear equations because we have three unknowns and three conditions.

We are required to find the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10)

The general quadratic function can be written as

$\displaystyle f(x)=ax^2+bx+c$

We need to find the values of a,b, and c. Let's use the first condition, i.e. f(-1)=1

$\displaystyle f(-1)=a(-1)^2+b(-1)+c$

$\displaystyle f(-1)=a-b+c$

$\displaystyle a-b+c=1.....[eq\ 1]$

Now we use the second condition f(1)=5

$\displaystyle f(1)=a(1)^2+b(1)+c$

$\displaystyle f(1)=a+b+c$

$\displaystyle a+b+c=5.......[eq\ 2]$

Finally, we use the third condition f(2)=10

$\displaystyle f(2)=a(2)^2+b(2)+c$

$\displaystyle f(2)=4a+2b+c$

$\displaystyle 4a+2b+c=10....[eq\ 3]$

We put together eq 1, eq 2, and eq 3 to form the system

$\displaystyle \left\{\begin{matrix}a-b+c=1\\ a+b+c=5\\ 4a+2b+c=10\end{matrix}\right.$

Adding the first two equations we have

$\displaystyle 2a+2c=6$

$\displaystyle a+c=3$

And also

$\displaystyle b=2$

Using the above equation and the value of b in the third equation, we have

$\displaystyle \left\{\begin{matrix}a+c=3\\ 4a+c=6\end{matrix}\right.$

Subtracting the first equation from the second

$\displaystyle 3a=3$

$\displaystyle a=1$

And therefore

$\displaystyle c=2$

Now we have all the values, the quadratic function is

$\displaystyle \boxed{f(x)=x^2+2x+2}$

6 0

3 years ago