Draw a shape with 4 sides and corners that is not a rectangle. explain your drawing.

Mathematics

2 answers:

lutik1710 [3]3 years ago

8 0

It's a square, parallelogram, or rhombus, because all three have 4 sides and corners and none of them are rectangles.

bija089 [108]3 years ago

4 0

Its a square that's the only other thing it could be ur welcome fammm

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If a(x) = 3x + 1 and b(x)= squareroot x-4, what is the domain of (b*a)(x)?

oksano4ka [1.4K]

$\boxed{ \ x \geq 1 \ }$ or can be written as $\boxed{ \ [1, \infty) \ }$

<h3>Further explanation</h3>

This is a question about the composition of functions and how to get a domain function.

Given $\boxed{ \ a(x) = 3x + 1 \ }$ and $\boxed{ \ b(x) = \sqrt{x - 4} \ }$ .

We will form (b o a)(x) and then determine the domain.

$\boxed{ \ (b \circ a)(x) = b(a(x)) \ }$

Replace each appearance of x in b(x) with $\boxed{ \ a(x) = 3x + 1 \ }$ .

$\boxed{ \ (b \circ a)(x) = \sqrt{(3x + 1) - 4} \ }$

Thus, $\boxed{ \ (b \circ a)(x) = \sqrt{3x - 3} \ }$

To be defined, the value under the radical sign must not be negative. Therefore, the domain of $(b \circ a)(x) = \sqrt{3x - 3}$ are processed as follows.

$\boxed{ \ 3x - 3 \geq 0 \ }$

Both sides added by 3.

$\boxed{ \ 3x \geq 3 \ }$

Both sides divided by 3.

$\boxed{ \ x\geq 1 \ }$

Thus, the domain of $(b \circ a)(x) = \sqrt{3x - 3}$ is $\boxed{ \ x \geq 1 \ }$ or can be written as $\boxed{ \ [1, \infty) \ }$

<h3>Learn more</h3>

If f(x) = x² – 2x and g(x) = 6x + 4, for which value of x does (f o g)(x) = 0? brainly.com/question/1774827
Solve for the value of the function composition brainly.com/question/2142762
Look for rotation rules in the transformation brainly.com/question/2992432

Keywords: composition of function, if a(x) = 3x + 1, and, b(x) = √(x-4), what is the domain of, (b o a)(x), b(a(x)), defined, the value, under the radical sign, must not be negative,