$5= x^(2) +2x$ solve the equation using any method you prefer. Give imaginary roots.

Answers

Answer 1

Answer: $5=x^2+2x\nx^2+2x-5=0\nx^2+2x+1-6=0\n(x+1)^2=6\nx+1=\sqrt6 \vee x+1=-\sqrt6\nx=-1+\sqrt6 \vee x=-1-\sqrt6$

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Solve. 10/3x + 4/3 = 7+x/2x
A) X = 1/3
B) X = 17/5x
C) X = 1/5
D) X = 1/6

Answers

well multiply both sides by 6x and you get 20+ 8x = 21 + 3x. so x=1/5 so (C)

Given the preimage and image, find the dilation scale factor

Answers

Answer:

1/2

Step-by-step explanation:

the dilation factor should be 1/2

if u count the dots on the sides of the preimage there are 2 dots that's the length
if u count the dots on the real image it's 4
therefore 2/4 to it's lowest term is 1/2
so 1/2 is the scale factor

Answer:

should be 1/2 as your answer

Write a set of parametric equations for y=x^2+1 given the parameter t=x-2

Answers

$t=x-2\ \ \ \Rightarrow\ \ \ x=t+2\n \ny=x^2+1\ \ \ \Rightarrow\ \ \ y=(t+2)^2+1=t^2+4t+5\n \n \left \{ {{x=t+2\ \ \ \ \ } \atop {y=t^2+4t+5}} \right.$

which of the following group of numbers are all prime numbers a.2,5,15,19 b.7,17,29,49 c.3,11,23,31 d.2,3,5,9,

Answers

Prime numbers are those that cannot be divided by another number (other than themselves and 1). In other words, prime numbers have factors of themselves and 1.

For example, 3 is a prime number because its only factors are 3 and 1.

6 is not a prime number because it can be factored to 3*2.

Therefore, the correct answer is c. 3,11,23,31

Prime numbers can only be divided by 1 and itself.

First group:- It contains 15, and 15 can be divided by 3 and 5 to get a whole number.

Second group:- 49 can be divided by 7 to receive 7. So this is wrong.

Third group:- This group can be considered all prime numbers, so this is correct.

Fourth group:- Incorrect. Because 9/3 = 3

Answer:- c. 3,11,23,31 are all prime numbers.

What is the perimeter of a tennis court 78ft long and 27ft wide?

Answers

210^2 is the answer L*W

Of the 1,000 students in a local college, 420 own brand X mobile phones and 580 own brand Y mobile phones. Of these students, 80 own both brands of mobile phones. Find the probability that a student chosen at random has a brand X mobile phone given that he has a brand Y mobile phone.2/14

5/21

3/28

4/29

Answers

Answer: The probability that a student chosen at random has a brand X mobile phone given that he has a brand Y mobile phone is 4/29.

Step-by-step explanation:

Since, the total number of students, n(s) = 1,000

The number of students who have X mobile phones, n(X) = 420,

And, number of students who have Y mobile phones, n(Y) = 580,

Thus, the probability of the student that has Y phones,

$P(Y)=(n(Y))/(n(S))=(580)/(1000)=0.58$

While, the number of students who have both phones, n(X∩Y) = 80

Thus, the probability of the student who has both phones,

$P(X\cap Y)=(n(X\cap Y))/(n(S))=(80)/(1000)=0.08$

Hence, the probability that a student chosen at random has a brand X mobile phone given that he has a brand Y mobile phone.

$P((X)/(Y))=(P(X\cap Y))/(P(Y))$

$=(0.08)/(0.58)$

$=(8)/(58)=(4)/(29)$

Hence, the required probability is 4/29.

If we were to put this in terms of a venn diagram, we would have 360 owning only brand X, 500 owning only brand Y, and 80 in between, owning both. Therefore, 80 out of the 580 owners of brand Y may have X as well, which we put into fraction form 80/580, and reduce to 4/29.

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