Simplify 2(5+3x) + (x+10)

How do I show the relationship between the input and the output

alexgriva [62]

Answer:

Types of Relationships between the Input and Output

The scatter plot can be a useful tool in understanding the type of relationship that exist between the inputs (X’s) and the outputs (Y’s)

Step-by-step explanation:

1. No Relationship: The scatter plot can give an obvious suggestion if the inputs and outputs on the graph are not related. The points will be scattered throughout the graph with no particular pattern. For no relationship to exist, points have to be completely diffused. If some points are in concentration, then maybe a relationship does exist and our analysis has not been able to uncover it.

2. Linear and Non-Linear: A linear correlation exists when all the points are plotted close together. They form a distinct line. On the other hand points could be close together but they could form a relationship which has curves in it. The nature of the relationship has wide ranging implications.

3. Positive and Negative: A positive relationship between the inputs and the outputs is one wherein more of one input leads to more of an output. This is also known as a direct relationship.

On the other hand a negative relationship is one where more of one input leads to less of another output. This is also known as an inverse relationship.

4. Strong and Weak: The strength of the correlation is tested by how closely the data fits the shape. For instance if all the points are scattered very close together to form a very visible line then the relationship is strongly linear. On the other hand, if the relationship does not so obviously fit the shape then the relationship is weak.

I don't know if this was exactly what you were looking for; hope it is! :)

7 0

3 years ago

Christian wants to make an A in his class. There are 5 exams this year, and he has taken 4 of them. His grades were 84, 88, 95,

ioda

Answer:

last test grade=x x ≥ 79 so 79%

Step-by-step explanation:

(84/5)+(88/5)+(95/5)+(104/5)+(x/5) ≥ 90

Simplify: (371/5)+(x/5) ≥ 90

Subtract (371/5) from both sides

Then you have (x/5) ≥ (79/5)

Then multiply both sides by 5 and you have x ≥ 79

P.S. If it was helpful please consider giving me brainliest. =)

3 0

3 years ago

In Japan. traditionally it was believed that if you folded 1,000 origami cranes, your wish would come true. For a party, Emika i

vfiekz [6]

Answer:

Step-by-step explanation:

Remark

Interesting detail.

Origami Paper is square -- perfectly.

So the properties listed will be related to a square.

Givens

Area = 35 in^2

Formulas

Area = s^2 where s is the length of 1 side.

Perimeter = 4s

Solution

Area = s^2 = 35

s^2 = 35 Take the square root of both sides.

√s^2 = √35

s = 5.91

Perimeter = 4s

s = 5.91

Perimeter = 4*5.91

Perimeter = 23.66

The way the question is worded, the answer you should submit is

Perimeter = 4 * √35

4 0

2 years ago

The perimeter of equilateral triangle ABC is 81/3 centimeters, find the length of the radius and apothem.

MAXImum [283]

There is a typo error, the perimeter of equilateral triangle ABC is 81/√3 centimeters.

Answer:

Radius = OB= 27 cm

Apothem = 13.5 cm

A diagram is attached for reference.

Step-by-step explanation:

Given,

The perimeter of equilateral triangle ABC is 81/√3 centimeters.

Substituting this in the formula of perimeter of equilateral triangle = $3\times\ side$

$3\times\ side$ $=[tex]81\sqrt{3}$

$Side = \frac{81\sqrt{3} }{3} =27\sqrt{3} \ cm$

Thus from the diagram , Side $AB=BC=AC= 27\sqrt{3} \ cm$

We know each angle of an equilateral triangle is 60°.

From the diagram, OB is an angle bisector.

Thus $\angle OBC = 30$ °

Apothem is the line segment from the mid point of any side to the center the equilateral triangle.

Therefore considering ΔOBE, and applying tan function.

$tan\theta =\frac{perpendicular}{base} \\tan\theta=\frac{OE}{BE} \\tan\theta=\frac{OE}{\frac{27\sqrt{3}}{2} } \\tan30\times {\frac{27\sqrt{3} }{2} }= OE\\\frac{1}{\sqrt{3} } \times\frac{27\sqrt{3} }{2} =OE\\$

Thus ,apothem OE= 13.5 cm

Now for radius,

We consider ΔOBE

$cos\theta=\frac{base}{hypotenuse} \\cos30= \frac{BE}{OB} \\Cos30 = \frac{\frac{27\sqrt{3} }{2}}{OB} \\OB= \frac{\frac{27\sqrt{3} }{2}}{cos30} \\OB= \frac{\frac{27\sqrt{3} }{2}}{\frac{\sqrt{3} }{2} } \\OB =27 \ cm$