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4 years ago

Taylor sorts 6 shapes in two groups based on attributes

Mathematics

2 answers:

lesya [120]4 years ago

5 0

Irregular hexagons (Meaning 6 sides, irregular is optional) and quadrilaterals.

FinnZ [79.3K]4 years ago

3 0

As seen from the figure:

Group 1: It consists of three closed figures each having six sides. So these are all hexagons.

Group 2: It consists of three closed figures each having four sides.

All four are quadrilaterals.

So we can say that group 1 consists of hexagons and group two consists of quadrilaterals.

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Answer:

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Step-by-step explanation:

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3 years ago

- f:R → R<br> f(x + 7) = 3x - 6<br> g(2x + 1) = x2-1<br> = (f-1og)(5) = ?

Murljashka [212]

Answer:

$f^{-1}\,\circ \,g(5) = 10$

Step-by-step explanation:

Let $f(x+7) = 3\cdot x -6$ and $g(2\cdot x +1) = x^{2}-1$ , we proceed to derive $f(x)$ and $g(x)$ by algebraic means:

(i) $f(x+7) = 3\cdot x -6$

1) $f(x+7) = 3\cdot x -6$ Given

2) $f(x+7) = 3\cdot (x+0) - 6$ Modulative property

3) $f(x+7) = 3\cdot [(x+7) +(-7)]-6$ Existence of additive inverse/Associative property

4) $f(x+7) = 3\cdot (x+7) +3\cdot (-7)-6$ Distributive property

5) $f(x+7) = 3\cdot (x+7) -21-6$ $a\cdot (-b) = -a\cdot b$

6) $f(x+7) = 3\cdot (x+7) -27$ Definition of subtraction

7) $f(x) = 3\cdot x - 27$ Composition of functions/Result

(ii) $g(2\cdot x + 1) = x^{2}-1$

1) $g(2\cdot x + 1) = x^{2}-1$ Given

2) $g(2\cdot x + 1) = (x\cdot 1)^{2}-1$ Modulative property

3) $g(2\cdot x +1) = [(2\cdot x)\cdot 2^{-1}]^{2}-1$ Existence of additive inverse/Commutative and associative properties

4) $g(2\cdot x +1) = (2\cdot x)^{2}\cdot 2^{-2}-1$ $a^{c}\cdot b^{c}$ / $(a^{b})^{c} = a^{b\cdot c}$

5) $g(2\cdot x + 1) = \frac{(2\cdot x)^{2}}{4}-1$ Definitions of division and power

6) $g(2\cdot x + 1) = \frac{(2\cdot x + 0)^{2}}{4} -1$ Modulative property

7) $g(2\cdot x +1) = \frac{[(2\cdot x + 1)+(-1)]^{2}}{4} -1$ Existence of additive inverse/Associative property

8) $g(2\cdot x + 1) = \frac{(2\cdot x + 1)^{2}+2\cdot (2\cdot x + 1)\cdot (-1)+(-1)^{2}}{4} -1$ Perfect square trinomial

9) $g(2\cdot x + 1) = \frac{(2\cdot x + 1)^{2}}{4}+\frac{[2\cdot (-1)]\cdot (2\cdot x + 1)}{4} +\frac{(-1)^{2}}{4}-1$ Addition of homogeneous fractions.

10) $g(x) = \frac{x^{2}}{4}-\frac{2\cdot x}{4} + \frac{1}{4}-1$ Composition of functions/ $a\cdot (-b) = -a\cdot b$

11) $g(x) = \frac{x^{2}}{4}-\frac{x}{2}-\frac{3}{4}$ Definitions of division and subtraction/Result

Now we find the inverse of $f(x)$ :

1) $f = 3\cdot x - 27$ Given

2) $f + 27 = (3\cdot x - 27)+27$ Compatibility with addition

3) $f+ 27 = 3\cdot x +[27+(-27)]$ Definition of substraction/Commutative and associative properties

4) $f+27 = 3\cdot x$ Existence of additive inverse/Modulative property

5) $(f+27) \cdot 3^{-1} = (3\cdot 3^{-1})\cdot x$ Compatibility with multiplication/Commutative and associative properties

6) $(f+27)\cdot 3^{-1} = x$ Existence of multiplicative inverse/Modulative property

7) $f^{-1} (x) = \frac{x+27}{3}$ Symmetrical property/Notation/Result

Finally, we proceed to calculate $f^{-1}\,\circ \, g (5)$ :

1) $f^{-1} (x) = \frac{x+27}{3}$ , $g(x) = \frac{x^{2}}{4}-\frac{x}{2}-\frac{3}{4}$ Given

2) $f^{-1}\,\circ\, g(x) = \frac{\frac{x^{2}}{4}-\frac{x}{2}-\frac{3}{4}+27}{3}$ Composition of functions

3) $f^{-1}\,\circ \,g(5) = 10$ Result

3 0

3 years ago

Koran is spending $12 on games and rides at another carnival, where a game costs $0.25 and a ride costs $1.

Vlad [161]

Answer:

$0.25x+y=12$

Step-by-step explanation:

Let

y ----> the number of rides

x ---> the number of games

we know that

The number of rides multiplied by $1 plus the number of games multiplied by $0.25 must be equal to $12 (the dollar amount Koran is spending)

The equation that represent this situation is

$0.25x+y=12$

5 0

3 years ago