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2 years ago

What are four different ways of representing functions? how can you tell if a relationship is a function?

1 answer:

6 0

A function can be represented verbally (using words), numerically, visually and algebraically. A relation is usually defined if something -- usually designated as x -- has in influence on the output -- usually designated as y.

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A circle has diameter of 11cm

dlinn [17]

Since length of diagonal ( $Diagonal= 9.9cm$ ) is less than diameter of circle ( 11 cm ) , Therefore , the square will fit inside the circle without touching the edge of the circle.

Step-by-step explanation:

Here we have , A circle has diameter of 11 cm A square has side length of 7 cm . Use Pythagoras’ Theorem to show that the square will fit inside the circle without touching the edge of the circle . Let's find out:

We know the concept that for any square to fit inside the circle without touching the edge of circle , diagonal of square must be less than diameter of circle . Let's find out length of diagonal by using Pythagoras Theorem :

$Hypotenuse ^2 = Perpendicular^2+Base^2$

For a square , $Perpendicular = base = side$

⇒ $Diagonal^2 = 2(side)^2$

⇒ $Diagonal= \sqrt{2}(side)$

⇒ $Diagonal= \sqrt{2}(7)$

⇒ $Diagonal= 9.9cm$

Since length of diagonal ( $Diagonal= 9.9cm$ ) is less than diameter of circle ( 11 cm ) , Therefore , the square will fit inside the circle without ruching the edge of the circle.

5 0

2 years ago

Square $ABCD$ has area $200$. Point $E$ lies on side $\overline{BC}$. Points $F$ and $G$ are the midpoints of $\overline{AE}$ an

Charra [1.4K]

1. Consider square ABCD. You know that

$A_{ABCD}=AD^2=200,$

then

$AB=BC=CD=AD=10\sqrt{2}.$

2. Consider traiangle AED. F is mipoint of AE and G is midpoint of DE, then FG is midline of triangle AED. This means that

$FG=\dfrac{AD}{2}=\dfrac{10\sqrt{2} }{2}=5\sqrt{2}.$

3. Consider trapezoid BFGC. Its area is

$A_{BFGC}=\dfrac{FG+BC}{2}\cdot h,$ where h is the height of trapezoid and is equal to half of AB. Thus,

$A_{BFGC}=\dfrac{FG+BC}{2}\cdot \dfrac{AB}{2}=\dfrac{5\sqrt{2}+10\sqrt{2}}{2}\cdot \dfrac{10\sqrt{2}}{2}=75.$

$A_{BFGC}=A_{BFGE}+A_{EGC},\\A_{EGC}=A_{BFGC}-A_{BFGE}=75-34=41.$

5. Note that angles EGC and CGD are supplementary and

$\sin \angle CGD=\sin \angle EGC.$

Then

$A_{CGD}=\dfrac{1}{2}CG\cdot CD\cdot \sin \angle CGD=\dfrac{1}{2}CG\cdot EG\cdot \sin \angle CGE=A_{ACG}=41.$

Answer: $A_{CGD}=41.$

8 0

2 years ago

PLS HALP ME!!!!!! You have $6.50 to make copies. Each copy costs $0.45. Write and solve an inequality that represents the number

serious [3.7K]

Step-by-step explanation:

Variable x = number of copies

6.50 <= 0.45x

Divide both sides by 0.45

6.50 ÷ 0.45 <= 0.45x ÷ 0.45

14.4444.... <= x

No more than 14 copies

6.50 <= 0.45x

6.50 <= 0.45(14.444)

7 0

2 years ago

this past Sunday, the giants scored 9 less than twice the cowboys. the packers scored 14 points more than the giants. if the tea

Grace [21]

Let the score of cowboys is x
and giants make score 9 which is twice less than the cowboys score so
giants score will be = 2x -9
and packers scored 14 more than giants that is (2x - 9) + 14
now sum of their scores is equal to 81 it means:
x + (2x - 9) + (2x -9) + 14 = 81
x + 2x - 9 + 2x - 9 + 14 = 81
5x = 81 + 4
5x = 85
x = 17
packers scored = (2x - 9) + 14
= 2 (17) -9 + 14
=38 + 5 = 43 points

5 0

2 years ago

The number of bagels sold daily for two bakeries is shown in the table. Bakery A Bakery B 53 34 52 40 50 36 48 38 53 41 47 44 55

IrinaK [193]

You are given the number of bagels sold daily for bakeries A and B and are shown in the table above. Based on the above data, it is better to describe the centers of distribution in terms of the mean than the median. This is because all of the data from bakery A and B have values that are near the average and no data sample have possible outliers. For instance,

bakery A
mean (A) = [53 + 34 + 52 + 40 + 50 + 36 + 48 + 38]/8 = 43.88

Bakery B
mean (B) = [53 + 41 + 47 + 44 + 55 + 40 + 51 + 39]/8 = 46.25

Even if bakery A and B has the 38 and 39 as the lowest data respectively, they are still near the average data.

7 0

2 years ago