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2 years ago

Prove that if a quadrilateral has four equal sides and one right angle, then the quadrilateral is a square

1 answer:

3 0

So lets ABCD are the sides of the square and AB=CD, AC=BD and the Angle ABC = 90 so therefore by making a diagonal we can use the SAS or SSS congruency for two triangles, so we can prove two triangles are equal and that is why the shape is square.

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Erin writes the set of ordered pairs below. The set represents a function.

tino4ka555 [31]

Point A shows that she is incorrect.
With functions, you can perform the "straight line test through each point. If the line goes through both points, you know it's not a function.
If we plotted point A, it'd fail the straight line test because the given point (-6, 7) already has -6 as an x value.
Hope this helps!

7 0

2 years ago

For the function f(x)=x2-4, fins the value of f(x) when x=6

garri49 [273]

$f(x)=x^2-4\\f(6)=6^2-4\\f(6)=32$

$f(x)=x*2-4\\f(6)=6*2-4\\f(6)=8$

P.S. Hello from Russia

7 0

2 years ago

Read 2 more answers

If both diagonals of a quadrilateral bisect each other when is the quadrilateral a parallelogram?

slamgirl [31]

It is more than just a quadrilateral. In fact it is going to be hard to pick.

These facts suit a square, a rectangle, a rhombus, and a parallelogram. And the above statement is true, but maybe a little harder to prove than the converse of the statement, which is the usual one you find.

The converse is "If you have a parallelagram, the diagonals bisect each other."

You might think a trapezoid deserves some mention. The diagonals of a trapezoid do not bisect each other.

6 0

2 years ago

What is the exact length of each side?

Alla [95]

Answer:

$6a^{2} \sqrt{ab}$

Step-by-step explanation:

area of square = side x side

thus, each side = square root of area

$\sqrt{36a^5b}$ = 6 $\sqrt{a^{4} ab}$ = 6 $a^{2}$ $\sqrt{ab}$

6 0

2 years ago

Write three consecutive numbers starting with number n+1, where n is natural number. Find the sum of those numbers and prove tha

netineya [11]

Consecutive numbers differ by one, which means that if you start with a certain integer $x$ , you obtain the next integer by adding one.

So, if you start with $n+1$ , the three consecutive numbers are

$n+1,\quad (n+1)+1 = n+2,\quad (n+2)+1 = n+3$

The sum of these three numbers is

$(n+1) + (n+2) + (n+3) = 3n+6$

This number is divisible by 3 because you can write it as

$3n+6 = 3(n+2)$

Note that this is true in general: the sum of three consecutive numbers is always divisible by three, because it is three times the middle number.

5 0

2 years ago