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3 years ago

Is this a polygon? If it is not a polygon please say why.

Mathematics

2 answers:

Eva8 [605]3 years ago

8 0

Answer:

if you where trying to draw it straight then yes it is a polygon but if it is supposed to be rounded then no it is not.

Step-by-step explanation:

a plane figure with at least three straight sides and angles, and typically five or more.

elena-14-01-66 [18.8K]3 years ago

3 0

Answer:

technically yes if you make the lines straight and not overlapping and touching another line.

Step-by-step explanation:

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Solve 2x2 + 12x − 14 = 0 by completing the square.

Elenna [48]

Answer:

the solutions are 1 and -7

Step-by-step explanation:

first add 14 to both sides

2x²+12x=14

2(x²+6x)=14 now find half of 6x and square it, then add it to both sides

2(x²+6x+9)=14+18(9 was multiplied by 2)

no factor

2(x+3)²=32 divide both sides by 2

(x+3)²=16 square root both sides

x+3=+/- 4 subtract 3

x= 4-3=1

x= -4-3=-7

4 0

2 years ago

Help pls i have 5 minss, no links

KonstantinChe [14]

Answer:

3/25

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Factories A, B and C produce computers. Factory A produces 4 times as manycomputers as factory C, and factory B produces 7 times

Elina [12.6K]

Answer:

The probability is $P(A') = 0.485$

Step-by-step explanation:

Let assume that the number of computer produced by factory C is k = 1

So From the question we are told that

The number produced by factory A is 4k = 4

The number produced by factory B is 7k = 7

The probability of defective computers from A is $P(A) = 0.04$

The probability of defective computers from B is $P(B) = 0.02$

The probability of defective computers from C is $P(C) = 0.03$

Now the probability of factory A producing a defective computer out of the 4 computers produced is

$P(a) = 4 * P(A)$

substituting values

$P(a) = 4 * 0.04$

$P(a) = 0.16$

The probability of factory B producing a defective computer out of the 7 computers produced is

$P(b) = 7 * P(B)$

substituting values

$P(b) = 7 * 0.02$

$P(b) = 0.14$

The probability of factory C producing a defective computer out of the 1 computer produced is

$P(c) = 1 * P(C)$

substituting values

$P(c) = 1 * 0.03$

$P(b) = 0.03$

So the probability that the a computer produced from the three factory will be defective is

$P(t) = P(a) + P(b) + P(c)$

substituting values

$P(t) = 0.16 + 0.14 + 0.03$

$P(t) = 0.33$

Now the probability that the defective computer is produced from factory A is

$P(A') = \frac{P(a)}{P(t)}$

$P(A') = \frac{ 0.16}{0.33}$

$P(A') = 0.485$

3 0

3 years ago

I need to give you my math numbers and then solve the step by step

Alchen [17]

Answer:

tell me

Step-by-step explanation:

7 0

3 years ago

Will mark branliest!!

zepelin [54]

It’s A Have Ah Good Day

6 0

2 years ago

Read 2 more answers

Is this a polygon? If it is not a polygon please say why.​

Is this a polygon? If it is not a polygon please say why.